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ELEMENTAEY   AND 
CCXNSTEUCTIONAL    GEOMETEY 


ELEMENTARY 


AND 


Constructional  Geometry 


BY 


EDGAE  H.  NICHOLS,  A.B. 

OF  THE  BROWNE  AND  NICHOLS  SCHOOL,  CAMBRIDGE,  MASS. 
I 


OF  THE     ^ 

UNIVERSITY 

OF 

»£iUFORN\L 


NEW  YOKE 
LONGMANS,  GEEEN,  AND  CO. 

LONDON  AND  BOMBAY 
1896 


/_T3  /  i>  ^ 


Copyright.  1896 

BY 

LONGMANS.  GREEN.  AND  CO. 


TWOW  DIMCTOHY 

miMTINO  AND  BOOKBINDING  COMPANY 

NEW  YORK 


PREFACE 

This  book  is  designed  for  pupils  beginning  Geometry  at 
the  age  of  twelve,  or  even  younger.  It  is  based  upon  the 
author's  class-room  experience  with  young  boys  during  the 
last  twelve  years.  It  is  hoped  that  it  may  prove  an  aid  to 
those  teachers  who  do  not  want  a  text-book  except  as  a 
guide  to  the  pupil,  to  enable  him  to  review  the  principles 
developed  in  the  class  by  answering  for  himself  a  set  of 
questions  similar  to  those  asked  in  the  class-room,  and  by 
working  out  practical  problems  illustrative  of  the  princi- 
ples studied.  It  cannot  be  used  as  a  text-book  from  which 
lessons  are  assigned  in  advance  to  the  class.  Each  prin- 
ciple should  be  developed  by  the  teacher  and  the  class, 
working  together — the  principle  coming  at  the  end,  not  at 
the  beginning,  of  the  study — and  then  a  lesson  from  the 
book,  covering  essentially  the  same  ground,  should  be  as- 
signed. The  book  has  not  been  divided  into  lessons,  be- 
cause sometimes  a  single  paragraph  will  afford  material  for 
several  lessons,  and,  again,  several  pages  can  be  mastered  in 
one  lesson.  Each  teacher  should  be  free  to  develop  his 
subject  in  his  own  way  without  artificial  lines  of  division. 

The  main  object  of  the  first  year's  study  is  to  make  the 
pupils  perfectly  familiar  with  the  use  of  their  tools,  so  that 
in  the  study  of  Theoretical  Geometry  the  construction  of 
the  figures  will  present  no  difficulty.  In  the  pursuit  of 
this  object  a  great  many  useful  working  principles  are 
learned,  which  greatly  simplify  the  more  advanced  study 
of  the  subject. 

The  proof  of  principles  is  introduced  very  gradually, 


183979 


VI  PREFACE 

and  the  use  made  of  even  the  few  proofs  given  must  de- 
pend upon  the  age  and  the  ability  of  the  particular  class 
concerned. 

From  the.  beginning  it  is  insisted  that  the  pupil  .shall 
have  a  clear  idea  of  the  exact  direction  of  any  line  named 
with  two  letters.  No  slovenliness  of  thought  should  be 
permitted  in  this  matter.  For  greater  accuracy  of  expres- 
sion two  words  have  been  coined — symparalhl  and  anti- 
parallel.  Parallel  lines  that  have  the  same  direction 
are  called  symparallel  lines,  and  those  that  have  exactly 
opposite  directions  are  called  antiparallel  lines.  Thus, 
antiparallel  lines  form  an  angle  of  180°  with  each  other, 
but  symparallel  lines  form  no  angle.  It  is  found  that  this 
distinction  prevents  all  confusion  in  the  study  of  parallels, 
and  helps  to  make  the  pupils  think  clearly  and  express 
themselves  with  accuracy.  Simple  and  natural  signs  are 
used  for  these  words. 

Attention  is  called  to  the  treatment  of  areas.  The  au- 
thor has  found  that  the  subject  lends  itself  especially  to 
the  object  aimed  at  in  the  first  year's  study.  An  indefinite 
field  for  the  ingenuity  of  the  pupils  is  opened,  and  often 
excellent  results  have  been  obtained.  In  his  manipulation 
of  areas  the  ordinary  problems  of  construction  (often 
reserved  as  the  very  last  subject  of  Geometry)  become  a 
mere  matter  of  course  to  the  pupil,  and  his  familiar  ac- 
quaintance with  valuable  principles  is  assured.  A  similar 
treatment  of  volumes  opens  a  still  larger  field,  if  the  time 
allotted  to  the  subject  allows  it. 

A  proper  use  of  the  Summary  will  add  greatly  to  the  in- 
terest and  to  the  value  of  the  subject.  In  the  first  place, 
each  teacher  can  word  his  definitions  to  suit  his  own  con- 
victions, and  can  decide  what  principles  he  will  take  up, 
and  how  those  principles  shall  be  expressed.  In  the  sec- 
ond place,  the  pupil  feels  an  added  interest  in  the  princi- 


PREFACE  vii 

pies,  because  he  has  helped  to  decide  upon  the  best  word- 
ing of  them.  An  important  feature  of  the  work,  there- 
fore, is  the  training  that  it  gives  in  accuracy  of  expression. 
The  teacher  ought  first  to  make  the  pupils  express  the  prin- 
ciples in  their  own  words,  and  then  by  interpreting  their 
English  literally,  to  bring  out  any  inaccuracies  of  expres- 
sion. 

A  pupil  who  has  acquired  a  familiar  knowledge  of  the 
principles  developed  in  this  book  should  be  able  to  take  up 
the  study  of  Theoretical  Geometry,  both  plane  and  solid, 
in  a  text-book  where  no  complete  proofs,  but  suggestions 
only,  are  given  to  aid  in  the  solution  of  the  more  difficult 
problems  and  theorems.  He  should  be  encouraged  to  con- 
sult, after  his  own  solution,  the  text-books  that  contain 
proofs  (of  which  there  should  be  several  kinds  in  the  school 
library),  in  order  to  develop  his  critical  power  and  to  add 
life  to  the  subject. 

The  author  has  in  mind  the  preparation  of  a  Geometry 
to  supplement  this  book,  in  which  the  principles  of  Plane 
and  Solid  Geometry  will  be  developed  according  to  the  sug- 
gestions just  made.  He  will  be  grateful  to  any  who  will 
call  his  attention  to  typographical  errors  and  ambiguous 
expressions  in  this  book  that  may  have  crept  in,  in  spite  of 
the  greatest  care  on  the  part  of  himself  and  of  friends,  who 
have  kindly  read  and  re-read  the  proofs.  He  is  indebted 
to  one  of  his  former  pupils  for  the  best  of  the  diagrams. 

Edgar  H.  Nichols. 

December,  1895. 


CONTENTS 


Preface 

EXERCISES 

PAGE 
V 

Section  I. 

Introductory    . 

1-23 

1 

Section  II. 

Introductory    . 

24-58 

5 

Section  III. 

Lines  .... 

59-74 

10 

Section  IV. 

Angles 

75-112 

14 

Section  V. 

Parallels . 

113-121 

22 

Section  VI. 

Review 

.       122-139 

26 

Section  VII. 

Parallels  (Continued) 

140-171 

28 

Section  VIII. 

Triangles  . 

172-195 

32 

Section  IX. 

Review 

196-209 

39 

Section  X. 

Isosceles  Triangles 

.       210-246 

41 

Section  XI. 

Triangles  (Continued) 

Section  XII. 

Quadrilaterals 

281-323 

55 

Section  XIII. 

Division  of  Lines 

324-366 

65 

Section  XIV. 

Multiplication  of  Lines  ;  Area 

i      367-398 

72 

Section  XV. 

Moulding  Areas  ;   Equivalen- 

p 

Figures  .... 

399-436 

79 

Section  XVI. 

Peculiarities  of  Squares 

87 

Section  XVII. 

Areas  of  Similar  Figures 

.       533-565 

104 

Section  XVIII 

Circles  and  Inscribed  Angles 

}      566-625 

111 

Section  XIX. 

Volumes     .... 

.       626-678 

123 

Index     . 

.  137 

Summary  of  Facts,  Definitions,  and  Prd 

(CIPLl 

2S 

.  139 

ELEMENT  AEY 

AND 

CONSTRUCTIONAL  GEOMETRY 

SECTION  I. 

1.  In  the  group  of  drawings  on  the  third  page  are  out- 
lines of  several  familiar  bodies  or  solids.  Learn  their 
names,  and  find  as  many  examples  as  you  can  of  each  kind 
of  solid  in  the  objects  about  you.  Can  you  name  any 
other  solids  not  represented  there  ? 

2.  One  of  the  aims  that  we  shall  keep  before  us  in  our 
first  study  of  Geometry  will  be  to  discover  the  characteris- 
tics of  these  and  other  common  solids.  We  shall  study  at 
first  with  special  care  the  nature  of  their  bounding  sur- 
faces and  of  their  bounding  lines ;  we  shall  learn  how  to 
make  accurate  drawings  and  measurements  of  them ;  and 
we  shall  study  the  peculiarities  of  form  in  the  lines,  sur- 
faces, and  solids. 

3.  In  looking  at  the  physical  solids  about  us,  we  think 
not  only  of  their  form,  but  also  of  the  material  of  which 
they  are  made,  of  their  color,  and  of  their  fineness  of  fin- 
ish ;  but  in  dealing  with  geometrical  solids,  we  think  of 
the  form  only.  If,  for  example,  you  move  a  cube  of  glass 
from  its  place,  and  then  imagine  the  space  which  it  first 
occupied  to  be  bounded  by  surfaces  and  lines,  it  will  help 


2  ELEMENTARY  AND 

you  to  form  an  idea  of  a  geometrical  solid.  You  must, 
however,  be  on  your  guard  against  one  danger.  Your 
glass  cube,  however  skilfully  made,  will  have  imperfec- 
tions in  it ;  the  lines  will  not  be  true,  or  the  surfaces  will 
not  be  perfectly  smooth  :  but  the  geometrical  cube  of  your 
imagination  is  absolutely  perfect ;  every  line  is  true,  every 
surface  perfectly  smooth.  It  is  a  great  aid  to  the  imagina- 
tion to  use  physical  solids  for  illustration ;  but  you  must 
never  forget  that  the  geometrical  solids  which  they  repre- 
sent, and  which  you  are  to  study,  have  none  of  their  im- 
perfections. In  all  your  drawings,  too,  you  must  aim  to 
have  your  work  as  near  perfection  as  is  possible ;  to  this 
end  you  must  provide  yourselves  with  well-sharpened  hard 
pencils,  a  good  ruler,  and  a  strong  pair  of  compasses.  It  is 
better  to  have  one  pencil  set  aside  for  drawing  only,  and  to 
have  that  sharpened  with  a  chisel  edge  for  use  with  a  ruler. 
But  you  must  never  imagine  that  the  solids  have  any  of  the 
faults  of  your  drawings. 

4.  The  geometrical  solid,  then,  is  a  portion  of  space 
bounded  by  imaginary  surfaces.  It  is  limited  in  three  di- 
rections at  right  angles  to  each  other ;  or,  in  other  words, 
it  is  said  to  have  three  dimensions,  length,  breadth,  and 
thickness.  Sometimes  height  or  depth  is  used  in  place  of 
thickness.  Can  you  think  of  any  practical  examples  where 
height  or  depth  would  be  a  better  word  than  thickness  ? 
Do  you  know  of  any  word  to  take  the  place  of  breadth  ? 

5.  Examine  a  cube  (Fig.  1).  Which  dimension  do  you 
call  its  length,  which  its  breadth,  and  which  its  thickness  ? 
Is  there  any  difference  in  the  magnitude  of  these  dimen- 
sions ?    Can  a  cube  have  unequal  dimensions  ? 

6.  Can  you  draw  a  cone  or  a  cylinder  with  equal  dimen- 
sions ? 

7.  What  do  you  consider  to  be  the  height  of  a  pyramid  ? 

8.  Can  you  suggest  any  practical  way  of  finding  the 
height  of  a  (physical)  pyramid  ? 

9.  Which  of  the  solids  represented  must  have  equal  di- 
mensions from  their  very  nature  ? 


CONSTRUCTIONAL   GEOMETRY  3 


Fig.  1. 


Fig.  2. 


Fig.  3. 


Fig.  4. 


Fig.  5.  Fig.  6. 


Fro.  7. 


Fig.  8. 


4  ELEMENTARY  AND 

10.  Is  there  any  solid  in  the  group  which  cannot  have 
equal  dimensions  without  changing  its  nature  ? 

11.  How  would  you  proceed  to  measure  the  dimensions 
of  a  cone  of  wood  ? 

12.  The  boundaries  of  a  solid  are  called  surfaces,  and  the 
intersections  of  the  surfaces  are  called  lines.  The  bound- 
ing surfaces  form  the  faces  of  the  solid  ;  the  intersections 
of  the  faces  form  the  edges  of  the  solid. 

13.  How  many  faces  are  needed  to  bound  a  cube  ?  How 
many  edges  ?  How  many  faces  and  edges  are  needed  for  a 
triangular  prism  ?  for  a  cone  ?  for  a  cylinder  ?  for  a  sphere  ? 

14.  How  many  and  what  dimensions  has  a  surface  ?  Is 
the  leaf  of  your  book  a  surface  or  a  solid  ?  Why  ?  Is  one 
surface  any  thinner  than  another  in  Geometry  ?  Is  a  coat 
of  paint  a  surface  P     Why  ? 

15.  Can  you,  by  placing  a  great  many  surfaces  together, 
one  upon  another,  form  a  solid  ?  Is  it  right  to  say  that  a 
surface  is  a  very  thin  piece  of  a  solid  ? 

16.  How  many  and  what  dimensions  has  a  geometrical 
line  ?  You  speak  of  drawing  a  line.  Is  what  you  draw  a 
geometrical  line  ?  What  is  it  ?  Why  ?  What  is  the  proper 
name  in  Geometry  for  a  "telegraph  line"  ?  for  a  pencil 
mark  ?  When  you  speak  of  a  "fine  line" or  of  a  "coarse 
line,"  what  are  you  really  describing  ? 

17.  The  ends  of  a  line  are  points.  What  dimensions  can 
a  point  have  ?  Can  you  make  a  point  with  a  pencil  ? 
When  you  indicate  the  position  of  a  point  by  a  pencil  mark 
or  dot,  what  is  the  proper  geometrical  name  for  the  dot  ? 

18.  How  many  points  are  indicated  in  Fig.  1  ?  (page  3) 
in  Fig.  3  ?  in  Fig.  4  ? 

10.  By  means  of  what  yon  have  already  learned,  complete 
the  sentences  that  follow,  and  copy  them  neatly  on  the 
blank  leaf  at  the  end  of  the  book  headed  "  Summary  of 
Facts,  Definitions,  and  Principles, "  or  copy  them  on  some 
special  page  of  your  note-book  as  yon  may  be  directed. 

20.  Solids  have dimensions,  and    are    bounded 

by 


CONSTRUCTIONAL  GEOMETRY  5 

21.  Surfaces  have dimensions,  and  are  bounded 

by 

22.  Lines  have dimensions,   and   are    bounded 

by ••• 

23.  Points  have dimensions ;  they  show  position 

only. 

SECTION  II. 

24.  If  it  is  not  possible  to  make  points,  or  to  draw  lines, 
are  we  not  blocked  at  the  very  beginning  of  our  course  ? 
Not  so,  if  we  remember  what  our  drawings  really  amount 
to  ;  they  are  merely  aids  to  the  imagination  ;  the  dot,  which 
stands  for  a  point,  enables  us  to  be  sure  that  all  are  think- 
ing of  the  same  point  or  position  ;  the  line  A  B  is  an  aid  to 
the  eye,  and  makes  it  easier  to  remember  that  a  line  is 
imagined  between  the  points  A  and  B ;  but  the  imagina- 
tion should  always  go  behind  the  poor  imitation  of  a  line 
to  the  absolutely  perfect  line  represented  by  it.  If  the  line 
is  not  needed  to  aid  the  imagination,  it  is  better  not  to 
draw  it ;  for  instance,  in  the  figures  of  Section  I.  a  great 
many  lines  are  left  for  the  imagination  to  supply,  because 
the  mind  does  not  need  them  to  form  a  true  picture.  Give 
examples  of  omitted  lines  in  those  figures.  With  the  ex- 
planation just  given  we  can 
speak  of  drawing  lines  and 
of  making  points  without 
causing  any  confusion. 

25.  Imagine  a  point  A 
to  move  from  its  position  to 
the  position  B.  It  is  clear 
that  you  can  imagine  it 
to  move  along  several 
paths.  How  many  ?  How 
many  dimensions  will  any 
one  of  these  paths  have  ?  Why?  What,  then,  is  the  proper 
name  for  the  imaginary  path  ?   See  20-23. 


6  ELEMENTARY  AND 

26.  We  can,  therefore,  describe  a  line  in  another  way, 
which  is  better  for  our  purposes  upon  the  whole,  as  fol- 
lows :  A  line  is  the  path  traced  by  a  point  as  it  moves 
from  one  position  to  another.  Record  this  definition  in 
the  u  Summary." 

27.  There  are  four  paths  or  lines  between  A  and  B  pict- 
ured in  the  figure  ;  the  upper  path  is  called  a  curved  line ; 
the  next,  A  C  D  B,  is  called  a  broken  line;  the  next,  A  B, 
a  straight  line  ;  and  the  bottom  one  a  mixed  line. 

28.  Write  out  a  definition  of  each  kind  in  your  own 
language,  aiming  to  deeoribe  the  characteristics  of  each  so 
clearly  that  there  will  be  no  confusion  among  the  kinds. 
Bring  your  results  into  the  class-room,  and  after  deciding 
upon  the  best  wording,  learn  the  definitions  which  ere 
selected  as  the  best.  After  this,  if  a  line  between  two 
points  is  mentioned,  a  straight  line  is  always  to  be  imagined 
unless  otherwise  gpecifled. 

'.".».  Select  examples  of  the  different  kinds  of  lines  from 
the  figures  1  to  8  on  page  :*. 

30.  Can  you,  without  removing  your  pencil  from  the 
pftper,  and  without  going  over  the  same  path  twice,  trace 
the  outline  of  Figs.  1,  2,  3,  5,  and 

31.  Can  you  trace  all  four  paths  in  Fig.  9  with  a  con- 
tinuous movement  ? 

:;.'.  With  your  compasses 
trace  a  circumference.  Write 
as  good  a  definition  as  you  can 
of  a  circumference  viewed  as 
the  path  of  a  moving  point. 

Draw  three  concentric  cir- 
cumferences. (Look  up  the 
meaning  of  the  word  concen- 
tric, if  necessary.) 

33.  Construct  Fig.  10  with 
your    compasses,   beginning  Pio.  10. 

with  the  central  circumference  and  using  one  radius  for 
all  the  circumference-.     What  is  meant  by  the  radius  of  a 


CONSTRUCTIONAL   GEOMETRY  7 

circumference  ?     Take  great  pains  with  this  figure,  and  do 
not  give  up  until  you  have  an  exact  and  neat  copy  of  it. 

34.  Construct  Fig.  11,  beginning  with  three  points  one 
inch  from  each  other. 


Fig.  11. 


Fig.  12. 


35.  Construct  Fig.  12,  putting  the  centres  at  the  corners 
of  a  square  an  inch  each  way. 

36.  Construct  a  spiral  (Fig.  13)  by  using  0  as  the  centre  for 
the  first  semi- circumference 
A  C  B  and  every  alternate 
one,  and  by  using  A  as  the 
centre  for  the  second  semi- 
circumference  B  D  E  and 
every  alternate  one. 

37.  Construct  Fig.  14. 

38.  Construct  Fig.  15,  by 
first  placing  two  drawing-pins 
or  small  tacks  about  2J  in. 
apart  at  M  and  N ;  then, 
fastening  a  piece  of  fine 
thread  3  in.  long  to  the  two  pins,  with  your  pencil  pressed 
firmly  against  the  thread,  trace  the  curve.  This  curve  is 
called  an  ellipse;  Try  the  experiment  of  putting  the  pins 
nearer  together  without  altering  the  length  of  the  string. 


Fig.  la 


OF  THE 

UNIVERSITY 


8 


KLKM E\TMIY  A XI) 


Put  the  pins  close  together ;  what  kind  of  a  curve  do  yon 
now  trace  ?  The  ellipse,  in  addition  to  being  a  beautiful 
curve,  has  a  peculiar  interest  to  us  because  it  is  the  path 


Fio.  15. 


Fio.  14. 


which  the  earth  traces  as  it  moves  about  the  sun.    The  sun 
is  at  one  of  the  places  represented  by  the  pins. 

39.  Imagine  a  point  to  move  from  A  to  D  (Fig.  16), 
tracing  the  line  A  D,  and  then  imagine  the  line  A  D  to 
move,  without  tipping,  to  a 
new  position,  B  C.  Can  the 
line  take  more  than  one  path  f 
How  many  ?  Make  illustra- 
tions for  your  answers. 

40.  How  many  dimensions 
has  any  one  of  these  paths  ? 
Why  ?  What  should  the  path 
be  called?    See  20-23. 

41.  Can  you  move  a  line  in 
such  a  way  that  it  will  not  generate  a  surface  ? 

42.  Can  you  trace  the  lateral,  or  side,  surfaces  of  Fig. 
1  by  the  movement  of  a  line  as  described  in  40  ?  of  Fig.  2  ? 
of  Fig.  4  ?  of  Fig.  5  ? 

43.  How  would  you  move  a  line  to  generate  the  lateral 
surface  of  Fig.  3  ? 


Fio.  16. 


CONSTRUCTIONAL  GEOMETRY 


9 


A 


Fig.  17. 


44.  What  kind  of  line  would  you  move  to  generate  the 
surface  of  Fig.  6,  and  in  what  way  would  you  move  it  ? 

45.  Can  you  generate  the  lateral  surface  of  Fig.  5  by 
the  movement  of  a  curved  line  ? 

46.  Draw  a  curved  line  ABGD  (Fig.  17).     Move  a  line 
A  P,  without  tipping  it,  so  that 

A  shall  remain  on  the  curve. 
Will  a  surface  or  a  solid  be  gen- 
erated ?    Why  ? 

47.  Compare  the  surfaces 
which  have  been  generated  in 
exercises  39-42,  and  tell  which 
are  plane  and  which  are  curved 
surfaces.  Write  a  good  defini- 
tion of  a  plane  surface  that  will 
distinguish  it  from  a  curved 
surface.  Select  the  best  definition  from  those  handed  in 
by  the  class,  and  record  it  in  the  "  Summary." 

Suggestion  :  Think  how  a  carpenter  tests  with  his  straight- 
edge the  evenness  of  a  floor  or  table. 

48.  Imagine  the  point  A  to  move  to  D  (Fig.  18),  generat- 
ing the  line  A  D  ;  the  line  A  D 
to  move  to  B  C,  generating  the 
surface  A  B  C  D  ;  and  the  sur- 
face A  B  C  D  to  move  to  the 
position  E  F  G  H.  Could  the 
surface  take  more  than  one  path 
in  moving  to  its  new  position  ? 
Illustrate  your  answer.  How 
many  dimensions  in  any  one  of 
the  paths  ? 

49.  Could    you    generate  by 
the  motion  of  a  surface  the  solids  of  Figs.  1,  5,  6,  and  8  ? 

50.  In  what  way  could  you  move  a  surface  to  generate 
the  solids  of  Figs.  3,  6,  and  7  ? 

51.  In  what  two  ways  could  you  generate  the  cylinder, 
Fig.  5  ? 


10  ELEMENTARY  AND 

52.  Can  you  move  a  surface  so  that  it  will  not  generate 
a  solid  P 

53.  Review  from  24,  and  complete  the  sentences  that  fol- 
low, following  the  directions  given  in  19  : 

54.  A  line  is  the  path  traced 

55.  A  straight  line  il  the  path  traced 

56.  A  surface  is  the  path  traced 

57.  A  solid  is  generated  when 

58.  A  solid  of  revolution  is  generated  when 


SECTION  III.    LINES. 

59.  Since  a  straight  line  is  merely  the  path  traced  by  a 
point  as  it  moves,  without  change  of  direction,  from  one 
position  to  another,  the  line  can  have  only  two  character- 
istics, which  are  determined  by  the  direction  in  which  the 
point  moves  and  by  the  distance  which  it  moves.  It  is 
v.  iv  important  in  naming  a  line  to  indicate,  if  possible,  the 
direction  of  it  ;  it  is,  therefore,  important  to  put  first  the 
point  from  which  the  moving  point  is  imagined  to  start; 

\ample,   the  line  A  B    .  .  Q 

(  Kg.  19)  is  the  path  traced   ' 

by  a  point  moving,  without 

change  of  direct  ion.  from  A  P 

to  B,  not  from  B  to  A.  N 

The  direction  of  the  line         u 
A  B  is,  therefore,  from  A  to 
B  :  the  length  of  A  B  is  the  *»• u 

distance  between  A  and  B.  Describe  the  lines  M  N,  M  A, 
\  M.  B  M  (Fig.  19)  in  a  similar  way.  It  ought  not  to  be 
necessary  for  you  to  represent  the  lines  in  this  case  to  get  a 
clear  idea  of  them.  The  points  that  limit  the  lines  should 
enable  you  to  form  a  distinct  picture  of  the  lines  in  your 
mind. 

60.  In  what  wavs  does  the  line  B  A  differ  from  I»  M  ': 


CONSTRUCTIONAL   GEOMETRY  11 

How  does  M  N  differ  from  MNP?  (Fig.  19).     How  does 
M  N  differ  from  N  M  ? 

61.  How  many  different  straight  lines  can  you  draw  from 
M/oN  ?  (Fig.  19).    Can  you  give  a  reason  for  your  answer  ? 

An  axiom  is  a  truth  that  is  assumed  as  self-evident,  so 
that  it  needs  no  reason.  Can  you  state  the  axiom  that  you 
have  made  use  of  in  your  answer  to  61  ? 

62.  Can  you  draw  through  M  and  N  more  than  one 
straight  line  ?  (The  straight  lines  that  you  draw  must 
differ  in  some  way. )  Can  you  draw  between  M  and  N  as 
the  limiting  points  more  than  one  straight  line  ?  Compare 
line  MNP  with  M  N,  and  N  M  with  M  N  before  giving 
your  answer,  and  be  sure  that  the  axiom  of  61  is  worded 
exactly. 

63.  Sometimes,  when  the  direction  of  a  line  is  of  no  con- 
sequence, a  line  may  be  named  by  a  single  letter;  for 
example,  A  B  (Fig.  19)  might  be  called  "  I"  (Small  letters 
are  used  when  it  is  desired  to  name  a  line  in  this  way.)  It 
is  clear,  however,  that  the  name  "  I '* gives  no  suggestion  as 
to  whether  the  line  points  from  A  to  B,  or  from  B/oA; 
that  is,  there  is  doubt  about  one  of  the  two  characteristics 
that  are  possible  for  a  line.  Therefore,  unless  the  direc- 
tion is  of  no  consequence,  or  unless  the  figure  in  itself  makes 
doubt  impossible,  it  is  unwise  to  use  a  single  letter  in  naming 
a  line.  Of  course  a  line  should  never  be  named  B  A,  if  A  B 
is  the  line  in  one's  mind.  Beginners  cannot  be  too  careful 
about  this  point. 

64.  On  one  indefinite  straight  line  mark  two  lines  that 
shall  differ  in  length,  but  not  in  direction  ;  also  on  one 
straight  line  mark  two  lines  that  shall  differ  in  direction, 
but  not  in  length. 

65.  On  one  straight  line  mark  two  lines  that  differ  both 
in  length  and  in  direction. 

66.  In  the  exercises  that  follow  pay  no  attention  to  the 
direction  of  the  lines,  but  try  to  estimate  accurately  the 
lengths  of  the  lines.  Estimate  the  shorter  lines  in  inches 
or  centimeters,   the  longer  lines  in  feet  or  decimeters. 


12  ELEMENTARY  AND 

Make  your  estimate  first,  and  record  it ;  then  measure  the 
line,  and  record  the  result  in  a  second  column  ;  in  i  third 
column  put  the  error  made. 

Make  five  lines,  a,  b,  c,  d,  e,  of  different  lengths  and  in 
different  positions  ;  estimate  their  lengths  in  inches,  and 
record  results  as  explained  above. 

G7.  Make  a  line,  and  try  to  make  five  others,  in  various 
positions,  of  the  same  length  ;  record  your  errors  in  milli- 
meters. 

68.  Make  a  six-sided  figure  with  sides  varying  consider- 
ably in  length.  Estimate  the  lengths  in  centimeters,  and 
record  estimated  lengths,  true  lengths,  and  errors  in  three 
columns  as  before. 

Let  one  of  the  class  put  ten  different  lines  on  the 
board,  keeping  to  himself  the  true  lengths  of  the  lines ; 
let  the  rest  of  the  class  hand  in  to  him  their  estimates  in 
feet,  and  let  him  average  the  errors  made  by  each  one. 

7<>.  Make  a  horizontal  line  ;  try  to  make  a  vertical  line 
and  a  slanting  or  oblique  line  of  the  same  length  as  the 
horizontal  line.  BeoOfd  errors  in  millimeters.  It  is  best 
to  measure  a  line  with  your  compasses,  and  to  apply  your 
compasses  to  the  ruler. 

71.  Estimate  in  inches  the  height  of  a  chair  (to  the 
seat),  the  dimensions  of  a  window-pane,  the  width  of  a 
door.     Record  results  as  before. 

Note. — Exercises  of  a  similar  nature  can  be  given  prof- 
itably, until  each  pupil  becomes  fairly  expert  in  estimating 
lengths. 

72.  In  comparing  the  lengths  of  two  lines,  a  and  b,  we 
say  that  a  is  equal  to  b,  a  is  greater  than  b,  or  a  is  less  than 
b,  according  to  the  facts,  recording  the  result  in  one  of  the 
following  ways  :  a  =  b,  a  >  b  or  a  <  b.  Note  the  meaning 
of  the  signs,  which  are  used  in  comparing  any  two  magni- 
tudes. 

Make  two  broken  lines  of  three  parts  each  ;  measure  the 
total  length  of  eacli  line,  and  put  the  proper  sign  of  the 
three  signs  just  explained  between  the  two  results. 


CONSTRUCTIONAL  GEOMETRY  13 

73.  Make  three  different  lines  a,  b,  and  c ;  make  one 
line  equal  to  a  +  b  +  c.  Make  another  line  equal  to  a  + 
b  —  c ;  another  equal  to  a  —  b  +  c.  Put  the  proper  sign 
between  a  +  b  —  c  and  a  —  b  +  c. 

74.  Making  use  of  the  lines  a,  b,  c,  d,  and  e  in  Fig.  20, 
find  the  line  equal  to  a  —  b  +  c  —  d  +e;  find  also  b  —  2c 


Fig.  20. 

+  d  —  3d ;  a  —  3c  +  b  —  e.  Record  the  length  of  the 
answer  in  millimeters  for  each  case.  Since  all  of  the  class 
are  using  the  same  lines,  the  answers  should  come  out  alike. 
Do  they  ?  If  not,  can  you  account  satisfactorily  for  the 
differences,  supposing  each  one  to  have  taken  the  greatest 
possible  care  ? 

Although  measuring  a  line  with  compasses  and  ruler 
seems  a  very  simple  thing  in  theory,  in  practice  it  is  a  very 
hard  thing  to  do,  when  great  accuracy  is  required.  This 
is  due  partly  to  the  fact  that  the  points,  which  represent 
the  ends  of  the  lines,  are  very  far  from  being  true  points, 
and  partly  to  the  inaccuracy  of  our  tools.  Two  lines  that 
seem  to  be  of  the  same  length,  when  measured  with  an 
ordinary  ruler,  might  prove  to  differ  distinctly  when 
measured  with  a  more  delicate  tool  such  as  the  micrometer 
screw.  Do  you  know  what  a  micrometer  screw  is  ?  If  not, 
let  some  of  the  class  find  out  what  it  is,  for  the  next  lesson, 
securing  one,  if  possible,  for  actual  use.  Do  you  know  of 
any  other  device  for  measuring  lengths  with  great  ac- 
curacy ? 

The  result  of  your  work  in  the  exercises  of  74  and  what 
has  just  been  said  should  teach  you  that  it  is  not  safe 
to  rely  wholly  upon  measurement  when  it  is  to  be  de- 
cided whether  two  lines  are  of  equal  length  or  not.     Lines 


14 


ELEMENTARY  AND 


that  seem  to  the  eye  or  to  the  tools  to  be  equal  are  often 
unequal  ;  while  lines  that  seem  to  be  unequal  arc  often 
equal.  The  sides  of  a  true  square,  for  instance,  we  know 
must  be  equal  from  the  nature  of  the  square,  but  the  sides 
of  the  figure  that  we  draw  to  represent  the  square  may  be 
unequal.  You  must,  of  course,  use  your  tools  to  draw 
figures  to  aid  your  imagination  ;  but  you  must  think  more 
of  the  imaginary  figures  than  of  the  actual  figure  W0D  by 
you,  and  you  must  not  think  it  a  sufficient  reason  to  prove 
two  lines  equal  in  length  to  say  that  you  have  measured  tin- 
lines  and  found  them  to  be  equal.  Some  reason,  based 
upon  the  nature  of  the  figure  as  in  the  case  of  the  square 
above,  must  be  given,  if  possible. 


SKCTION    IV.     AXGLKS. 


75.  In  the  last  few  exercises  you  have  thought  only  of 
the  length  of  the  lines ;  now  consider  their  direction. 
If  two  lines  do  not  point  exactly  alike,  they  are  said  to 
make  an  angle  with  each  other.  Thus  in  the  oblong 
ABC  I)  K  PG  (Fig.  W),  the 
edge  A  B  is  said  to  make  an 
angle  with  the  edges  AD,  B  C, 
B  I\  B  B,  and  D  G,  because  it 
does  not  point  as  they  du. 

Does  A   B  make    an    angle 
with  DO?     Why? 

Does  A  B  make  an  angle  with 
G  F  ?  with  C  B  ?    Why  ? 

You  have  very  likely  used  the 
word  "  angle  "  in  the  place  of  the  word 
in  saying  that  a  house  was  in  the  angle  of  two  streets,  you 
meant  that  it  was  just  at  the  corner  in  the  space  between 
the  two  streets  ;  or,  perhaps,  you  have  thought  of  an  angle 
as  the  Space  included  between  two  lines.     But  in  Geometry 


"corner;"  thus, 


CONSTRUCTIONAL  GEOMETRY 


15 


the  angle  has  to  do  solely  with  the  direction  of  the  lines  ; 
if  the  lines  point  alike,  they  form  no  angle  ;  if  they  differ 
in  direction,  they  form  an  angle.  To  make  an  angle  larger, 
we  do  not  produce  its  sides  that  they  may  inclose  more 
space  between  them,  but  we  make  the  difference  in  direc- 
tion greater  by  turning  one  of  the  lines  away  from  the 
other.  Starting  with  your  compasses  closed,  turn  one  leg 
gradually  away  from  the  other :  the  two  legs  at  once 
begin  to  point  in  different 
directions  so  that  they  form 
an  angle  with  each  other  ; 
as  you  keep  on  turning, 
the  legs  point  farther  and 
farther  away  from  each 
other,  going  through  the 
various  positions  shown  in 
Fig.  22,  and  the  angle  grows 
larger  and  larger.    In  time  Fig.  22. 

the  turning  leg  points  exactly  opposite  to  0  A  in  the  direc- 
tion 0  F.  If  we  turn  the  leg  still  farther  to  the  position  of 
0  G,  there  is  not  so  much  difference  in  the  way  the  legs 
point  as  there  was  when  the  turning  leg  was  at  0  F  ;  in 
other  words,  the  angle  is  growing  smaller  again  ;  if  we  keep 
on  turning,  presently  the  leg  will  point  again  as  0  A  does, 
and  there  will  be  no  angle  formed. 

76.  Enough  has  now  been  said  about  an  angle  to  enable 
us  to  say  what  the  angle  really  is.  An  angle  is  the  differ- 
ence in  direction  of  two  lines.  It  indicates  how  much  you 
must  turn  one  of  the  lines  to  make  it  point  like  the  other. 
The  amount  that  one  of  the  lines  has  to  be  turned  to  make 
it  point  like  the  other  is  a  very  good  measure  of  the  size  of 
the  angle.  The  angle  is  not  the  corner  or  the  place  where 
two  lines  meet ;  it  is  not  the  space  between  the  lines  ;  it 
is  not  anything  that  you  can  see  :  but  it  is  a  difference 
to  be  determined  by  the  judgment,  as  any  other  differ- 
ence is  determined.  Perhaps  the  best  illustration  from 
familiar  things,  to  help  us  to  form  a  clear  idea  of  what 


16  ELEMENTARY  AND 

an  angle  really  is,  is  to  be  found  in  colors  or  in  sounds. 
We  see  two  shades  of  red,  for  instauce.  We  see  that 
there  is  a  difference,  but  we  do  not  see  the  difference 
itself  ;  that  is  something  to  be  determined  by  judgment. 
If  we  are  well  trained  in  distinguishing  colors,  we  can 
estimate  the  difference  in  shade  with  great  accuracy. 
Again,  when  we  hear  two  different  notes  on  the  piano,  we 
can  estimate  the  difference  in  pitch  exactly,  if  our  ears 
have  been  trained  to  distinguish  sounds.  In  like  manner,  if 
we  train  ourselves  to  distinguish  differences  in  direction,  we 
can  estimate  the  size  of  the  angle  with  surprising  accuracy. 
We  must,  however,  have  some  means  of  expressing  our 
estimates.  In  colors  ire  estimate  by  shades;  in  sounds,  by 
tones  and  semi-tones.  What  is  to  take  the  place  of  these 
measures  in  the  case  of  angles  ? 

77.  In  your  experiment  in  75  (Fig.  22),  as  you  turned 
the  leg  of  the  dividers,  you  made  0  H  point  more  and  more 
away  from  the  direction  of  0  A,  but  nearer  and  nearer  to 
the  direction  of  0  F,  which  has  the  direction  of  AO  exactly 
opposite  to  that  of  OA.  There  must  have  been  one 
position  of  O  B  whin  it  diverged  from  0  A  just  as  much  as 
it  did  from  OF.  In  that  position  (0  D  in  Fig.  88)  it 
made  a  right  angle  with  0  A  and  also  with  0  F,  so  that 
the  adjacent  ingiM  POD  and  AOD  were  equal.  In 
Geometry  the  right  angle,  or  the  angle  formed  when  one 
line  meets  another  so  as  to  make  the  adjacent  angles  equal. 
serves  to  divide  angles  into  two  large  classes.  Any  angle 
less  than  a  right  angle  is  called  an  acnte  angle  ;  any  angle 
larger  than  a  right  angle  is  called  an  obtuse  angle.  It  is 
also  common  to  call  all  angles  that  are  not  right  angles, 
whether  acute  or  obtuse,  oblique  angles. 

78.  Select  as  many  right  angles  as  you  can  in  Fig.  21. 
In  an  ordinary  room  the  two  walls  intersect  in  a  corner  line 
which  makes  a  right  angle  with  the  edges  of  the  floor. 
Imagine  several  lines  on  the  floor  starting  from  a  corner. 
What  angles  do  these  lines  make  with  the  corner  line  of  the 
two  walls  ?    Can  you  draw  on  the  floor  or  ceiling  a  straight 


CONSTRUCTIONAL   GEOMETRY  17 

line  that  does  not  make  a  right  angle  with  this  corner 
line  ? 

The  right  angle  is  perhaps  the  most  interesting  of  all  the 
angles.  It  is  the  angle  that  the  carpenter  must  under- 
stand thoroughly;  if  he  makes  mistakes  in  his  right 
angles,  his  work  can  never  look  well.  How  many  devices 
of  the  carpenter  for  getting  true  right  angles  do  you  know  ? 

79.  Another  angle  that  plays  an  important  part  in 
Geometry  is  the  angle  formed  when  one  line  points  in  a 
direction  exactly  opposite  to  that  of  another  line.  This 
angle  is  called  a  straight  angle  ;  it  is  the  largest  angle  dealt 
with  in  elementary  Geometry  ;  it  indicates  that  one  of  the 
lines  would  have  to  make  a  half  revolution  to  point  like 
the  other.     Select  examples  of  straight  angles  in  the  room. 

80.  The  division  in  77  of  angles  into  two  large  classes  is 
not  definite  enough  for  our  purposes.  For  greater  definite- 
ness  in  estimating  angles,  a 
semi  -  circumference  is  di- 
vided into  one  hundred  and 
eighty  equal  parts  called  de- 
grees. The  number  of  de- 
grees that  0  A  (Fig.  23) 
would  have  to  pass  over  in 
reaching  the  position  of  0  0 
is  an  exact  measure  of  the  size  of  the  angle  between  0  A  and 
0  C.  If  it  is  found  that  0  A  would  have  to  turn  past  thirty 
divisions,  the  angle  is  said  to  be  an  angle  of  thirty  degrees- 
written  30°.  And  0  A  would  make  an  angle  of  30°  with 
any  line  pointing  exactly  like  0  C,  no  matter  where  in 
space  it  might  be. 

81.  How  many  degrees  in  every  right  angle  ?  How 
many  degrees  in  the  angle  between  0  A  and  OB?  (Fig.  23). 
How  many  degrees  must  there  be  in  any  angle  to  make  it 
obtuse  ?    How  many  degrees  in  a  straight  angle  ? 

82.  What  do  you  mean  by  saying  that  the  angle  between 
0  A  and  0  D  is  70°  ?  60°  ?  90°  ?  180°  ?  Answer  each 
part  in  a  full,  clear  sentence. 

2 


18 


ELEMENTARY  AND 


83.  The  instrument  used  for  measuring  angles  is  a 
graded  half  circle  of  metal  or  cardboard  called  a  protractor 
(Fig.  24).     The  centre  is  clearly  marked  by  a  notch.     The 


Wm  M 


inner  edge  is  graded  into  180  equal  parts  from  right  to  left ; 
the  outer  circle  is  graded  into  180  equal  parts  from  left  to 
right. 

84.  To  make  an  angle  of  70°  with  your  protractor  first 
draw  a  line  of  any  convenient  length  (M  N  Fig.  25),  then 
place  your  protractor  with  the*  cen- 
tre mark  at  M,  and  with  one  of  the 
zero  lines  along  M  N  :  next,  put  a 
dot  1*  at  the  point  exactly  opposite 
the  line  marked  70°  on  your  pro- 
tractor, remove  von  r  protractor,  and 
join  M  with  P.  If  you  wish  the 
line  to  be  above  M  N,  you  should 
use  the  70°  mark  of  the  inner  circle ; 
if  you  wish  it  to  be  below  M  N,  use 
the  outer  circle.     Draw  both  cases. 

85.  With  your  protractor  make  angles  of  22°,  60°, 
125°,  and  170°. 

8G.  Draw  two  lines,  A  IS  and  A  C  crossing  each  other  at  A 


M 


N 


Fig.  25. 


4.V 


Fig.  26. 


CONSTRUCTIONAL   GEOMETRY  19 

(Fig.  26).    Explain  fully,  step  by  step,  the  way  in  which  you 
would  use  your  protractor  to  measure  the  angle  between 
A  B  and  A  0.    Measure   it 
twice,  placing  the  zero  line 
of  your  protractor  first  on 
A  B  and  then  on  A  C. 

87.  For  the  sake  of  brev- 
ity the  following  signs  are 
used :  /  =  angle  ;  /  s  = 
angles;  /CABor/BAC 
±5  angle  between  A  B  and 
A  C.  In  using  the  last  sign 
be  careful  to  put  the  letter 
naming  the  point  at  which  the  two  lines  meet  between  the 
other  two.  The  angle  between  A  B  and  A  C  could  not  be 
written  /ABCor/ACB;  A,  the  meeting-point  of  the 
two  lines,  called  the  vertex  of  the  angle,  must  be  the  middle 
letter.  What  is  the  plural  of  vertex  ?  If  only  one  angle  is 
formed  at  a  vertex,  the  angle  may  be  named  by  the  vertex 
letter.  In  explaining  the  meaning  of  /  B  A  C,  you  must 
remember  to  read  the  line  from  the  vertex  out ;  for  in- 
stance, /  B  A  C  is  the  difference  in  direction  between  A  B 
and  A  0  ;  it  shows  how  much  you  must  turn  A  B  to  make 
it  point  like  A  0.  It  would  not  do  to  substitute  B  A  for 
A  B,  or  C  A  for  A  C.  If  you  think  for  a  moment  how 
much  you  would  have  to  turn  A  B  to  make  it  point  like 
C  A,  you  will  see  the  need  of  being  careful. 

88.  Explain  the  meaning  of  /  MJN  R,  /NMS,  /PTQ, 
without  drawing  the  lines  that  form  the  angles. 

89.  Make  any  figure  with  five  sides.  Estimate  the  angles 
of  the  figure  ;  record  your  estimate  in  one  column  ;  the 
true  value,  found  with  your  protractor,  in  another ;  and 
the  error  in  a  third  column. 

90.  Repeat  89,  with  a  four-sided  figure. 

91.  How  would  you  measure  the  angle  between  A  B  and 
C  A  (Fig.  26)  ? 

First  estimate  it  and  then  measure  it. 


L>0 


ELEMENTARY  AND 


92.  Estimate  the  value  of  the  angles  1,  2,  3,  4,  5,  6,  and 
7  in  Fig.  27,  recording  estimates,  true  values,  and  errors 
as  in  89. 


m.  2:. 


90°  free 
Repeat 


93.  Make  angles  of  60°,  20°,  110°,  45°,  and 
hand.     Test  your  accuracy  with  the  protractor. 
this  exercise,    drawing  lines 
that  do  not  meet. 

94.  Divide  an  angle  into 
two  equal  parts  with  your 
protractor. 

95.  Divide  an  angle  into 
three  equal  parts  with  your 
protractor. 

96.  Make  an  angle  CAB 
(Fig.  28)  with  the  aid  of  your 
protractor,  and   a  line   0  R. 

Make  an  angle  with  vertex  at  0  equal  to  /  CAB. 

97.  Make  a  five-sided  figure  or  pentagon.     Make  a  sec 
ond  pentagon  with  the  same  angles  hut  different  sides. 


CONSTRUCTIONAL  GEOMETRY  21 

98.  Make  two  angles  that  to  your  eye  seem  to  be  of  the 
same  size.  Can  you  test  their  equality  by  using  your  com- 
passes without  the  aid  of  your  protractor  ?  Explain  your 
steps  carefully. 

99.  Can  you  make  one  angle  equal  to  another  without 
your  protractor  ?     Give  three  illustrations. 

100.  Make  a  four-sided  figure,  or  quadrilateral,  and 
make  another  quadrilateral  with  different  sides  but  the 
same  angles,  using  your  compasses  and  ruler  only. 

101.  Make  two  angles,  A  B  C  and  D  E  F ;  make  an  angle 
that  you  think  is  equal  to  the  sum  of  /  A  B  C  and  /  D  E  F. 
Test  your  accuracy  with  the  help  of  your  compasses. 

102.  Can  you  explain  how  to  double  an  angle  ABC? 

103.  Can  you  add  two  angles  ?  three  angles  ?  Make 
your  work  clear  in  all  the  steps. 

104.  Draw  two  angles ;  add  them  with  the  aid  of  com- 
passes and  ruler  only;  test  your  result  by  finding  the 
number  of  degrees  in  each  angle  with  your  protractor. 

105.  Make  a  figure  with  three  sides,  called  a  triangle ; 
add  the  three  angles  formed  at  the  three  corners.  How 
many  degrees  are  there  in  the  angle  which  you  get  for  your 
answer  ? 

106.  Make  four  triangles  of  different  sizes  and  shapes. 
Add  together  the  angles  of  each  one,  and  note  the  number 
of  degrees  in  the  resulting  angle  each  time.  In  adding 
the  angles  use  ruler  and  compasses  only.  In  gotting  the 
number  of  degrees  use  a  protractor,  if  necessary.  Do  your 
experiments  seem  to  point  to  any  principle  ? 

107.  Cut  out  a  paper  triangle  ABC;  cut  off  the  cor- 
ners along  wavy  lines,  and  place  the  vertices  A,  B,  and 
C  together.  Does  this  experiment  confirm  the  previous 
experiments  ? 

108.  Make  a  quadrilateral,  and  add  the  angles  formed. 
Try  two  figures  differing  in  shape  and  size.  If  one  line 
should  turn  through  all  the  angles  of  the  quadrilateral,  how 
far  would  it  turn  ?  Measure  each  angle  with  a  protractor  : 
find  the  number  of  degrees  in  all  four  angles  of  each  figure. 


22  ELEMENTARY  AND 

109.  Make  a  five-sided  figure,  and  add  the  angles  formed. 

110.  Can  you  draw  any  conclusion  about  the  sum  of  the 
angles  of  a  figure  of  many  sides  ? 

111.  Verify  the  results  of  109  and  110,  by  cutting  off 
the  corners  of  paper  figures  of  the  corresponding  number 
of  sides.     Be  careful  to  mark  the  corners  before  cutting. 

112.  Can  you  describe  a  way  of  finding  the  difference  be- 
tween two  angles  with  compasses  and  ruler  only  P  Make 
two  angles,  and  find  an  angle  equal  to  their  difference. 


SECTION  V.     PARALLELS. 

113.  We  have  seen  that,  if  two  lines  point  in  the  same 
direction,  they  cannot  form  an  angle.      Do  you  remember 

why  P    We  have  seen  also  that,     

if  they  point  in  exactly  op-    A  B 

potite   directions,   A  B    and 
1 1  0,  for  example,  in  Fig.  29, 

they  form    a   straight  angle    

C  D 

Whether  two  lines  point  the  Fl°-  w 

same  way  or  exactly  opposite  ways,  they  are  said  to  be  paral- 
lel, because  they  run  b\j  the  side  of  each  other.  (The  name 
is  taken  from  two  Greek  words.)  It  is  a  great  convenience 
to  be  able  to  distinguish  the  two  cases.  When  two  lines  are 
parallel  and  point  the  same  why  the  following  sign,  f  f  .  will 
be  used,  and  the  lines  will  be  called  symparallel  lines.  For 
instance,  A  BffCDistobe  read  A  B  points  in  the  same 
direction  as  C  1),  or,  more  briefly,  A  B  is  symparallel  to  C  D. 
When  the  two  lines  are  parallel  and  point  in  opposite  direc- 
tions, the  sign  f  4  will  be  used,  and  the  lines  will  be  called 
antiparallel  lines.  Thus  ABfiDC  means  that  A  B  runs 
in  the  opposite  direction  to  1)  C,  or  A  B  is  antiparallel  to 
D  C.  If  single  letters  are  used,  the  direction  of  the  lines  is 
not  indicated ;  so  that  the  sign  without  the  arrow-heads 


CONSTRUCTIONAL   GEOMETRY  23 

can  be  used.  Thus  a  ||  b  mean  that  the  line  "a"  points 
either  in  the  same  direction  as  " b"  or  in  the  opposite  di- 
rection to  "  b." 

114.  Draw  a  line  A  B,  and  choose  a  point  C  anywhere 

outside  of  A  B  (Fig.  30).    Can    , 

you  draw  through  0  two  dis-  A  6 

tinct  lines  diverging  from  each 

other   and    yet   symparallel  to 

A  B  ?    It  does  not  take  long  to  C  • 

see  that  the  problem  is  impossi-  FlQ#  ^ 

ble,  but  when  we  try  to  give  a  reason  for  it,  we  find  that  it 

is  one  of  the  truths  that  is  self-evident  or  axiomatic.     It  is 

equally  self-evident  that  one  line  can  be  drawn  through  C 

in  the  plane  of  the  paper  pointing  like  A  B.     The  axiom 

can  be  stated  as  follows  :  Through  any  point  in  space  one 

line,  and  only  one  line,  can  be  drawn  in  any  desired  direction. 


Fig.  31a. 


115.  Imagine  two  lines,  A  B  and  C  D,  to  have  the  same 
direction  (Fig.  31a),  and  imagine  that  E  F  crosses  these 
lines  at  K  and  M  respectively.     At  once  we  see  the  possi- 


24  ELEMENTARY  AND 

bility  of  a  great  many  angles.  There  are  six  angles  with 
their  vertices  at  K,  and  six  with  their  vertices  at  M.  Can 
you  account  for  them  all  ? 

Can  you  see,  without  the  aid  of  your  protractor  or  of 
your  compasses,  that  some  of  the  angles  must  be  equal  from 
the  nature  of  the  case  ? 

Look  at  the  angles  EKB  and  E  M  D,  for  instance ; 
B  K  B  shows  you  how  much  you  must  turn  K  B  to 
make  it  point  like  K  K  ;  /  E  M  D  shows  you  how  much 
you  must  tuni  M  D  to  make  it  point  like  M  E.  Now  the 
lines  K  B  and  M  I)  point  in  the  same  way  at  the  start ; 
they  also  point  in  tin*  same  way  when  in  their  final  posi- 
tions, K  E  and  M  E.  Therefore  the  amount  of  turning 
has    been    the    sun.-    in    both    cases.    Or,     in    other    words, 

/  E  K  B  =  /_  KM  1).  Another  way  of  looking  at  the 
same  thing  is  this  :  since  the  direction  of  K  B  is  the  same 
as  that  of  M  D,  and  since  the  direction  of  K  E  is  the  same 
as  that  of  M  K,  the  ilificrrnce  in  direction  between  K  Band 
K  B  must  be  the  same  as  the  difference  in  direction  between 
M  D  and  M  E  ;  just  as  if  two  shades  of  white  are  the  MUM 
and  two  shades  of  red  are  the  same,  the  difference  between 
a  Bhade  of  the  white  and  a  shade  of  the  red  must  be  the  same 
in  one  case  as  in  the  other.  Our  knowledge  that  these  two 
angles  are  equal  belongs  to  the  kind  of  knowledge  referred 
to  in  the  last  part  of  74,  knowledge,  that  is,  which  is  based 
upon  the  nature  of  the  case,  and  which  is  not  affected  by 
defective  instruments  or  drawings.  Does  it  make  any  differ- 
ence in  the  equality  of  the  two  angles,  if  F  E  is  more  or  less 
inclined  to  the  parallel  lines  ? 

Make  the  following  principle  true  by  crossing  out  the 
words  that  need  to  be  crossed  out : 

Two  angles  are  or  are  not  necessarily  equal,  if  their 
sides  have  the  same  directions  respectively. 

Learn  the  principle  by  heart,  and  also  copy  it  in  the 
"  Summary." 

116.  Write  the  same  principle  almost  entirely  by  signs, 
which  have  already  been  explained. 


[    UNIVERSITY 

V  OF  J 

N^lifoRJJ^ 

^*^M==:^^VONSTRUCTIONAL  GEOMETRY  25 

117.  Select  other  pairs  of  angles  in  Fig.  31a  to  which  the 
principle  applies. 

118.  What  can  yon  say  about  the  angle  A  K  B  ?  What 
about  the  angle  E  K  M  ?  What  about  their  relation  to 
each  other  ?  Does  your  knowledge  of  the  facts  depend 
upon  measurements,  or  upon  the  nature  of  the  figure  ? 
Can  you  put  what  you  have  learned  in  this  exercise  into 
the  form  of  a  principle  ?  Select  for  the  "  Summary  "  the 
wording  finally  agreed  by  the  class  to  be  the  best. 

119.  /  B  K  M  is  one  part  of  a  straight  angle.  What  is 
the  other  part  ?  Two  answers  can  be  given  to  this  question. 
Can  the  answers  have  different  values  ?  Why  ?  When 
two  angles  together  make  a  straight  angle,  the  angles  are 
said  to  be  supplements  of  each  other.  Can  one  angle  be 
the  supplement  of  two  unequal  angles  ?  Why  ?  What  is 
the  supplement  of  75°  ?  of  110°  ?  of  82°  ?  of  60°  ?  Draw 
the  supplement  of  each  angle  of  Fig.  27. 

120.  There  is  an  axiom  that  if  equals  are  taken  from 
equals,  the  remainders  are  equals.  Do  you  understand  what 
it  means,  and  do  you  see  that  the  truth  has  anything  to  do 
with  exercise  119  ?  with  the  principle  of  115  ? 

121.  Select  in  Fig.  31a  as  many  pairs  of  supplementary 
angles  as  you  can.  Note  specially  those  which  are  the 
supplements  of  the  same  angle.  What  can  you  say  about 
the  direction  of  their  sides  ? 

Your  study  of  the  last  three  articles  should  enable  you 
to  make  the  following  statement  true :  Two  angles  are  or 
are  not  necessarily  equal  if  their  sides  have  opposite  direc- 
tions respectively.  Record  the  principle  in  the  "Sum- 
mary." Select  as  many  pairs  of  equal  angles  in  Fig.  31a  as 
you  can  that  are  equal  by  application  of  this  new  principle. 

If  A  K  M,  Fig.  31a,  is  25°,  find  all  the  other  angles  in 
the  figure. 

IfLAKM  =  |LMKB3  find  the  number  of  degrees 
in  all  the  angles. 

IfLBKM  =  4xLCMF,  find  the  number  of  degrees 
in  all  the  angles. 


26 


ELEMENTARY  AND 


Scholium.1  The  angles  that  are  formed  when  one  line 
crosses  two  other  lines  are  of 
such  great  importance  in 
Geometry  that  they  have  re- 
ceived special  names.  Those 
without  the  two  lines,  or  1,  2, 
7,  8  in  Fig.  31b,  are  called 
exterior  angles  ;  those  within 
the  two  lines,  or  3,  4,  5,  6,  are 
called  interior  angles  ;  I  and 
8  or  1  and  7  are  called  alter- 
nate-exterior angles ;  3  and 
5  or  4  and  <»  an-  called  alter- 
nate-interior angles ;  while  1 
and  5,  2  and  6,  3  and  7,  or  4  and  8,  are  called  correspond- 
ing angles. 


Fio   "lb. 


SECTION  VI.    REVIEW. 

122.  What  is  the  difference  between  a  geometrical  solid 
and  a  physical  solid  ? 

128.  What  is  a  plane  surface  ?  How  is  a  plane  surface 
tested  practically  t 

134,  How  many  kinds  of  lines  can  you  draw  between  two 
points  A  and  B  ?  Which  is  the  shortest  of  them  all  ?  How 
do  you  know  that  it  is  the  shortest  ?  How  does  the  straight 
line  A  B  point — from  B  to  A  ?  or  from  A  to  B  ? 

125.  Give  definitions  of  concentric  and  adjacent,  with 
illustrations  of  their  use  in  Geometry. 

186.  What  is  an  axiom  ?  Give  any  axioms  that  you  have 
learned. 

127.  Explain  how  you  use  a  protractor. 

128.  Write  in  sign  language :  A  B  is  longer  than  C  D. 


1 A  scholium  is  a  remark  containing  some  explanation. 


CONSTRUCTIONAL   GEOMETRY  27 

The  angle  A  B  C  is  less  than  the  angle  II  S  T.    A  B  points 
in  the  same  direction  as  C  D. 

129.  Write  in  good  English  the  following : 
/ABC=/EFG  because   B  A    f  f    F  E  and   B  C 

f  f  F  G,  or  because  their  sides  /  f  respectively. 

130.  What  is  a  pentagon  ?  a  quadrilateral  ?  a  triangle  ? 
What  is  a  straight  angle  ?  an  obtuse  angle  ? 

131.  Does  a  straight  line  change  its  direction  in  any  part 
of  its  course  ?  Can  you  place  two  points  so  that  you  can- 
not draw  a  straight  line  through  them  ? 

If  two  straight  lines  pass  through  the  same  two  points 
can  they  diverge  from  each  other  ?     See  113. 

132.  Can  you  think  of  a  way  of  testing  the  straightness 
of  your  ruler's  edge  by  two  points  on  your  paper  ? 

133.  What  is  an  angle  ?  Can  you  really  draw  an  angle  ? 
What  is  understood  by  ' '  drawing  an  angle"  ?  What  differ- 
ence in  the  size  of  the  angle  does  the  length  of  the  sides 
make  ? 

134.  Place  five  points  so  that  no  three  shall  be  on  the 
same  straight  line.  How  many  straight  lines  can  you 
draw  that  shall  each  contain  two  of  these  points  for  its 
ends  ?  Can  you  place  the  five  points  sp  that  only  four 
lines  can  be  drawn  each  containing  two,  and  only  two, 
points  of  the  five  as  the  limiting  points  ? 

135.  Make  as  clear  as  you  can  the  diiference  between  the 
straight  angle  ABC  and  the  straight  line  ABC. 

136.  What  is  the  vertex  of  the  angle  MPN,  and  what 
are  its  sides  ?    When  is  /  M  P  N  a  right  angle  ? 

137.  When  are  two  angles  supplements  of  each  other  ? 
Give  numerical  examples  and  geometrical  examples. 

138.  In  how  many  points  can  two  straight  lines  cross 
each  other  ?  two  circumferences  ?  In  how  many  points 
can  a  straight  line  cross  a  circumference  ? 

139.  In  what  cases  can  you  say  that  two  angles  are  equal, 
without  measurement  with  protractor  or  compasses  ? 
Write  out  the  principles  of  115  and  121,  making  use  of  the 
terms  explained  in  the  scholium  to  121. 


28  ELEMENTARY  AM) 


SECTION  VII.     PARALLELS  CONTINUED. 

140.  Draw  two  lines  A  B  and  A  C  (Fig.  32) ;  choose  any 
point  P  on  A  C.     How  many  lines  can  there  be  radiating 
from  I'  that  will  make  the  same 
angle  with   P  C  that  A  B  does 
with   A  Of     Answer  first   with 
the  understanding  that  the  line 
can  be  drawn  from  1'  in  any  di- 
rection  in   space ;  then   answer 
the  same  question  with  the  un- 
derstand in;:  t  hat   the  line  must 
be  drawn  on  the  same  plane  sur- 
face with  A  B  and  A  C,  for  in-  PiaW- 
stance,  on  the  plane  of  your  paper.      Actually  draw  the 
lines  in  the  second  case. 

141.  Draw  two  lines.  A  B  and  A  C,  as  in  Fig.  32.  At  the 
point  P.  wwh  the  aid  of  your  protractor,  make  an  angle 
equal  to  /  C  A  B,  drawing  your  second  line  P  R  so  that  it 
is  on  the  same  side  of  A  C  with  A  B.  Since  P  B  will  have 
to  diverge  from  P  C  just  as  much  as  A  B  does,  and  since 
l'K  and  A  B  are  on  the  same  side  of  A  C  and  in  the  same 
plane,  thev  must  point  in  the  same  direction,  or,  in  sign 
language,  if  /  ('  P  B  =  /  C  A  B,  P  R  f  f  A  B. 

1 1  .\  Could  two  lines  possibly  point  in  the  same  direc- 
tion and  not  be  in  the  same  plane  with  each  other  ?  See 
111. 

143.  Repeat  the  work  of  141,  using  your  compasses  instead 
of  your  protractor  to  make  equal  angles. 

144.  The  last  four  exercises  lead  to  the  following  impor- 
tant principle  :  If  two  lines  drawn  in  the  same  plane 
diverge  to  the  same  extent  from  one  side  of  a  third  line, 
they  point  in  the  same  direction,  or  are  symparallel. 
Make  clear  why  it  is  necessary  to  say  M  in  the  same  plane  " 
and  "  from  one  side/'  by  giving  illustrations  of  lines  that 


CONSTRUCTIONAL   GEOMETRY  29 

are  not  symparallel  because  they  fail  to  answer  all  the  con- 
ditions. Write  the  principle  given  above,  using  the  nota- 
tion explained  in  121,  scholium. 

145.  Make  a  line  A  B,  and  choose  any  point  0  outside  of 
the  line.  Draw  through  C  a  line  symparallel  to  A  B. 
Suggestion  :  Draw  C  A  as  a  help  line,  and  use  the  principle 
of  144. 

146.  Make  four  lines  symparallel  to  a  line  A  B  drawn 
at  pleasure.  Will  these  lines  be  symparallel  to  each 
other  ? 

147.  Make  a  three-sided  figure.  Through  each  corner 
draw  a  line  symparallel  to  the  opposite  side. 

148.  Make  a  quadrilateral  with  its  opposite  sides  parallel. 
Such  a  figure  is  called  a  parallelogram. 

149.  Can  you  discover  any  truth  about  the  angles  of  a 
parallelogram  without  actual  measurement  ?  Record  the 
principle  in  the  "  Summary. " 

150.  Make  a  triangle  ABC,  and  through  the  vertex  C 
(Fig  33)  draw  D  E  f  f  A  B.  Select  pairs  of  equal  angles, 
telling  in  each  case  how  you 
know  that  the  angles  are 
equal.  What  can  you  say 
about  the  sum  of  the  angles 
with  vertices  at  C  ?  What 
can  you  say  about  the  sum  of 
the  angles  of  the  triangle  ? 
Compare  the  result  obtained  Fia33. 

here  with  the  result  of  your  experiments  in  106.  Notice 
the  difference  in  the  methods  of  reaching  the  conclusion. 
In  this  case  the  conclusion  is  reached  after  studying  one 
triangle,  in  106  after  experimenting  with  several  triangles  ; 
in  this  case  what  we  know  about  the  angles  depends  upon 
the  nature  of  angles,  while  in  106  our  knowledge  is  based 
upon  our  measurements.  We  might  have  some  doubt  as  to 
whether  the  truths  learned  about  the  few  triangles  with 
which  we  experimented  in  106  would  apply  to  all  triangles  ; 
but  we  are  absolutely  sure  that  the  principle  discovered  in 


30  ELEMENTARY  AND 

150  applies  to  all  triangles,  because  it  is  based  on  the  nature 
of  angles  and  lines.  In  this  case  the  principle  is  proved ; 
in  106  it  is  merely  tested  by  experiment.  Record  the  prin- 
ciple in  the  "Summary." 

151.  Make  a  triangle  ABC  with  /  B  A  C  =  70°,  and 
I A  B  C  =  43°.  How  large  is  /  A  C  B  ?  Why  1  Through 
each  vertex,  A,  B,  and  C,  in  turn  draw  a  line  parallel  to 
the  opposite  side.  Four  new  triangles  will  thus  be  formed 
Give  the  value  of  each  angle  in  degrees  without  using  your 
protractor. 

152.  One  angle  of  a  triangle  is  35°,  another  47°  ;  find 
the  third  angle.  If  two  angles  of  a  triangle  are  48°  and 
95°  respectively,  what  is  the  third  angle  ? 

153.  Can  you  make  a  triangle  whose  angles  are  90°,  40°, 
and  60°  ?  Why  ?  Can  you  make  a  triangle  with  two  ob- 
tuse angles  in  it  }  Can  you  make  a  triangle  two  of  whose 
angles  shall  be  right  angles  ? 

1 54  Make  any  triangle.  Can  you  make  a  second  triangle 
which  shall  have  two  angles  like  two  of  the  first  triangle, 
but  which  shall  have  its  third  angle  unlike  the  third  angle 
of  the  first  triangle  ? 

155.  If  the  angles  of  Fig.  34  are  two  angles  of  a  triangle, 
how  can  you   find    the  third 

angle  without  using  your  pro- 
tractor ?  Can  you  think  of 
more  than  one  way  of  solving 
this  problem  ? 

156.  Two  angles  of  a  tri- 
angle are  said  to  determine  the 

third  angle.  Give  illustrations  of  the  meaning  of  this 
statement. 

157.  If  one  angle  of  a  parallelogram  is  equal  to  /  x  (Fig. 
34),  how  can  you  find  the  other  three  angles  without  con- 
structing the  parallelogram  ?    See  149. 

158.  One  angle  of  a  parallelogram  is  72°.  What  are  the 
other  angles  ? 

159.  Find  all  the  angles  of  a  parallelogram,  if  one  angle 


CONSTRUCTIONAL   GEOMETRY  31 

is  90°  ;   if  one  angle  is  120°  ;   if  one  angle  is  /  y  (Fig. 
34). 

160.  How  many  angles  of  a  parallelogram  are  required 
to  determine  the  remaining  angles  ?  Put  your  answer  in 
the  form  of  a  principle,  and  record  it  in  the  "  Sum- 
mary." 

161.  In  the  parallelograms  whose  angles  you  have  stud- 
ied thus  far,  can  you  discover  a  principle  that  applies 
always  to  the  adjacent  angles  ? 

162.  Extend  side  C  D  of  the  parallelogram  A  B  C  D  to 
R.  What  can  you  say  about  the  respective  directions  of 
the  sides  of  /  R  D  A  and  /  D  A  B  ?  What  follows  from 
the  directions  of  the  sides  of  the  angles  ? 

163.  Make  an  angle  ABC.  At  some  point  P  on  A  B 
betiveen  A  and  B  draw  (on  the  opposite  side  of  A  B  from 
B  C)  a  line,  P  R,  that  shall  diverge  from  P  B  as  much  as 
B  0  diverges  from  B  P.  What  is  the  geometrical  name 
for  angles  R  P  B  and  P  B  C  (121  Scholium)  ?  Which 
of  the  following  sentences  is  true  ?  P  R  ff  B  C  or 
PR/,/  B  C  ?  Write  a  translation  of  the  correct  sentence. 
Is  P  R  parallel  to  B  C  ? 

164.  Make  a  line  A  B  and  choose  a  point  C  outside  of 
A  B.  Draw  a  line  through  C  parallel  to  A  B,  without 
prolonging  your  help  line  A  C  beyond  C.  Will  the  line  be 
antiparallel  or  symparallel  to  A  B  ? 

165.  Repeat  the  problem  of  164  three  times,  placing  A, 
B,  and  0  in  different  positions  each  time. 

166.  The  last  four  exercises  lead  to  the  following  prin- 
ciple :  If  two  lines  are  crossed  by  a  third  in  such  a  way 
that  the  alternate  interior  or  alternate  exterior  angles  are 
equal,  the  two  lines  are  parallel. 

167.  Combining  the  principles  of  144  and  166,  we  can 
say  that  two  lines  are  parallel,  if  a  third  line  crosses  them 
in  such  a  way  as  to  make  any  pair  of  corresponding  angles 
equal,  or  any  pair  of  alternate  interior  or  alternate  exterior 
angles  equal. 

168.  A  practical  way  of  drawing  parallels  by  making  the 


32 


ELEMENTARY  AND 


corresponding  angles  equal  is  to  slide  a  triangular  piece  of 
wood  or  stiff  paper  along  a  ruler  until    the  edge  passes 
through    any   desired  point. 
For  instance,  in  Fig.   35    it 
is   desired   to   draw    a    line 
through  P   parallel  to  A  B. 
Place  your  ruler  in    such  a 
position  that  one  edge,  x,  of 
your  triangle  will   have   the 
direction  of  A  B  ;  then  slide 
the  triangle  along  the  ruler 
until  the  same  edge,  x,  passes 
through  P,  and  draw   M  N. 
The   ruler  represents  a   line 
crossing   M  N   and   A  B   in 
such  a  way  as  to  make  the  corresponding  angles  equal. 
1  ♦'.:♦.  Make  a  parallelogram  by  the  method  described  in 
168. 

170.  Mark  off  six  equal  distances  on  a  line  A  B  ;  then, 
by  the  method  of  108,  draw  parallels  through  the  points  of 
division. 

171.  Make  a  triangle  ;  then  through  each  vertex  draw  a 
line  parallel  to  the  opposite  side,  using  the  method  of  108. 


vie  m 


SECTION  VIII.     TRIANGLES. 


172.  Make  a  triangle  ABC  (Fig.  36).  The  sign  for  tri- 
angle is  l\,  and  for  triangles  ^.  What  is  the  difference 
between  /\ABC  and  /ABC?  Give  as  full  an  answer 
as  you  can  to  this  question,  naming  all  the  respects  in  which 
they  differ.     Are  they  alike  in  any  respect  ? 

173.  What  is  the  difference  between  £±  A  B  C  and 
/\BAC?  What  is  the  difference  between  /  A  B  C  and 
/  B  A  C  ?  In  how  many  ways  can  you  name  the  /\ABC, 


CONSTRUCTIONAL  GEOMETRY  33 

and  in  how  many  ways  can  you  name  the  /  A  B  C,  us- 
ing the  letters  A,  B,  C  each  time  ? 

174.  Write  a  short  sketch  of  the  A  A  B  c>  telling  how 
many  parts  it  has,  what  it  is  by  nature,  what  you  have 


Fig.  36. 


learned  about  its  angles,  and  what  fact  about  its  sides  you 
can  infer  from  124.  Also  write  all  you  can  about  /ABO. 
In  writing  these  sketches,  take  great  pains  with  your  Eng- 
lish. 

175.  After  making  A  A  B  C  (Fig.  36),  mark  off,  on  a 
long  line,  M  N  equal  in  length  to  A  B  ;  then  make  /NMR 
equal  to  /  B  A  C  ;  then  mark  off  on  M  R  a  length  M  P 
=  AC;  and,  finally,  join  P  N,  thus  completing  /\  M  N  P. 
Notice  that  you  made  the  lines  M  N  and  M  P  and  the 
angle  NMR  to  match  the  corresponding  parts  of  the 
/\  A  B  C,  but  that,  after  taking  these  steps,  you  had  no  con- 
trol over  the  line  P  N",  because  you  were  obliged  to  draw  it 
between  two  fixed  points  P  and  N.  With  your  protrac- 
tor compare  /  M  P  N  with  /  A  C  B,  and  /  M  N  P  with 
/ABC;  and,  with  your  compasses,  compare  P  N  with  C  B, 
recording  your  results  by  means  of  the  sign  =  or  the  sign 
4=  (unequal)  as  the  case  may  be.  If  you  find  that  P  N  is 
not  equal  to  C  B,  write  PN  ^  C  B,  but,  if  you  find  that 
P  N  is  equal  to  C  B,  write  P  N  =  C  B,  etc. 

Record  your  work  from  the  beginning  according  to  the 

model  which  follows,  taking  pains  to  record  each  step  as 

you  take  it,  not  waiting  to  give  your  description  after  all 

the  work  is  done.     Put  your  figure  at  the  upper  rig  Jit  cor- 

? 


34  ELEMENTARY  AND 

ner  of  your  page,  unless  it  is  so  large  that  it  requires  the 
whole  width  of  the  page. 

I.  I  make ^ABC.  Position  of  figure. 

II.  I  make  M  N  =  A  B. 

III.  I  make  /NMR=/BAC. 

IV.  I  make  M  P  =  A  C. 

V.  I  join  P  with  N,  thus  completing ^MNP. 
VI.  I  find  that  /  M  P  K  +  or  =  /  A  C  B  (using  the 

proper  sign). 
VII.  I  find  that  /  M  N  B  *  or  =  /  A  B  C. 
VIII.  I  find  that  P  N  *  or  =  C  B. 

t70.  Repeat  the  experiment  of  175  three  times,  using 
triangles  of  new  shapes  each  time,  and  recording  your  work 
according  to  the  above  model. 

177.  Do  your  experiments  in  175  and  176  seem  to  point 
to  any  truth  about  triangles  ?  Take  care  to  express  the 
principle  in  the  simplest  and  most  accurate  wording  pos- 
sible. 

178.  If  your  work  was  neatly  and  carefully  done  in  1  ;:> 
and  170,  you  probably  were  able  to  discover  the  principle 
involved.  But,  at  the  best,  you  have  merely  tested*  prin- 
ciple in  a  few  cases  ;  you  have  not  proved  it  by  showing 
that  it  depends  upon  the  nature  of  triangles,  or  upon  prin- 
ciples already  proved.  Let  us  try  to  prove  the  principle 
that  two  triangles  must  be  equal  in  all  respects,  if  they 
have  two  sides  and  the  included  angle  alike  respectively. 

There  is  a  very  great  difference  at  the  very  start  between 
testing  and  proving  a  principle.  In  testing  a  principle, 
everything  depends  upon  the  accuracy  of  the  mechanical 
work  ;  if  there  is  the  slightest  error  in  making  the  angles 
or  the  lines,  the  test  is  a  failure ;  but  in  proving  a  prin- 
ciple the  drawing  is  simply  to  aid  the  imagination,  by 
making  it  unnecessary  to  carry  a  great  many  lines  and  points 
in  the  mind  at  once  ;  the  accuracy  of  the  drawing  makes  no 
real  difference  in  the  results,  although  an  inaccurate  draw- 
ing often  misleads  the  mind ;   the  mind  should  be  con- 


CONSTRUCTIONAL  GEOMETRY  35 

stantly  on  the  imaginary  lines  and  angles  represented,  not 
on  the  lines  and  angles  drawn. 

Imagine,  then,  two  triangles,  ABO  and  M  N  P,  to  have 
two  sides,  A  B  and  A  C,  and  the  included  angle,  B  A  C,  of 
the  first  equal  respectively  to  two  sides,  M  N  and  M  P,  and 
the  included  angle  M  N  P  of  the  second.  You  are  to  try 
to  prove  that  the  remaining  parts,  B  C  and  the  angles  ABC 
and  A  C  B  of  the  first  triangle,  must  be  equal  to  the  re- 
maining parts,  N  P,  and  the  angles  MNP  and  MPNof 
the  second  triangle.  Imagine /\MNP placed  on/\ABC, 
and  try  to  prove  that  the  triangles  must  exactly  fit  one 
another.  How  many  points  can  you  place  before  /\MNP 
becomes  fixed  in  position  ?  After  placing  M  on  A,  are  you 
free  to  place  N  wherever  you  please  ?  In  how  many  places 
could  you  put  it  ?  Which  one  will  you  choose  ?  After 
placing  M  and  N  in  two  definite  positions,  in  how  many 
positions  could  you  place  P  ?  Remember  that  P  must  be 
somewhere  on  the  same  plane  surface  with  /\ABC,  since 
you  imagined  the  two  triangles  placed  together.  Have  you 
any  reason  for  thinking  that  P,  if  on  the  same  side  of  A  B 
with  C,  must  be  on  the  line  AC?  What  is  your  reason  ? 
If  you  can  prove  that  it  must  be  somewhere  on  A  C,  do  you 
know  exactly  where  it  must  be  ?  Remember  that  M  is 
fixed  at  A.  How  do  you  know  its  exact  position  on  A  C  ? 
Why  is  there  now  no  doubt  as  to  the  position  of  N  P  ?  See 
61,  Axiom.  If  you  can  find  satisfactory  answers  to  the 
questions  asked  above,  and  if  you  can  explain  your  answers 
by  reasons  that  depend  upon  the  nature  of  lines  and  angles 
or  upon  principles  already  proved  (and  you  ought  to  be 
able  to  do  this  with  ease),  you  have  proved  the  principle 
that  two  triangles  are  necessarily  equal  in  all  respects,  if 
they  have  two  sides  and  the  included  angle  alike  respec- 
tively. 

179.  By  the  aid  of  this  principle  of  178  we  are  able  to 
decide  without  measurement  that  two  lines  must  be  equal 
under  certain  circumstances,  and  we  have  found  a  new  way 
for  determining  the  equality  of  two  angles  without  protract- 


36  ELEMENTARY  AND 

or  or  compasses  ;  for  two  lines  or  two  angles  must  be  equal, 
if  they  are  corresponding  parts  of  triangles  that  are  known 
to  be  equal. 

180.  The  principle  of  178  is  of  great  practical  value  in 
enabling  us  to  find  the  distance  from  one  point  to  another 
that  cannot  be  seen  from  the  first  point  because  it  is  hid- 
den by  a  building,  or  wall,  or  some  intervening  obstacle. 
Suppose  you  wish  to  find  the  distance  from  a  stake  on  one 
side  of  a  house  to  a  stake  on  the  opposite  side  of  the  house. 
Represent  the  position  of  the  first  stake  by  a  point  A.  and 
that  of  the  second  stake  by  a  point  B.  Then  choose  a 
point  0  from  which  you  can  see  both  A  and  B,  and  measure 
the  angle  A  C  B. 

To  measure  the  angle  A  C  B,  you  need  an  enlarged  pro- 
tractor mounted  on  a  tripod.  If  possible,  get  some  sur- 
veyor to  show  you  his  theodolite,  and  to  explain  to  you 
how  lie  uses  it.  If  yon  divide  a  semicircle  having  a  radius 
of  four  or  five  inches  into  degrees,  and  mount  it  on  a  tripod 
such  as  you  would  use  for  a  camera,  you  can  measure  angles 
with  sufficient  accuracy  for  your  purposes.  A  stick  about 
a  foot  long,  provided  with  sights  and  pivoted  at  the  centre 
of  t he  protractor,  will  take  the  place  of  the  surveyor's  tele- 
scope. From  the  centre  of  the  protractor  suspend  a  plumb- 
line.  Placing  the  tripod  so  that  the  point  C  is  exactly 
below  the  centre  of  the  protractor,  sight  the  point  A,  and 
note  the  division  at  which  the  line  of  sight  crosses  the  pro- 
tractor ;  then  turn  the  stick  until  you  can  sight  point  B, 
and  again  note  the  division  of  the  protractor  crossed  by  the 
line  of  light  :  from  the  two  readings  of  the  protractor  esti- 
mate /  ACB.  By  "  sighting  point  A  "  is  meant  sighting 
a  staff  held  in  a  vertical  position  at  A.  Next,  measure  the 
distances  C  A  and  C  B.  If,  with  the  measurements  that 
you  have  made,  you  construct  a  second  triangle  in  a  level 
place  where  there  are  no  obstacles,  you  will  have  a  triangle 
exactly  like  /\  A  B  C,  so  that  you  can  measure  the  line  cor- 
responding to  A  B. 

181.  If  at  any  time  you  should  be  without  a  mounted 


CONSTRUCTIONAL  GEOMETRY  37 

protractor,  can  you  think  of  any  way  in  which  you  could 
use  a  piece  of  string  and  a  stake  to  measure  /  A  C  B  in 
the  last  problem  ? 

182.  Describe  the  measurements  that  you  would  take  to 
find  the  distance  across  a  pond. 

183.  In  practice  it  is  often  inconvenient  to  find  a  suita- 
ble place  in  which  to  make  a  triangle  as  large  as  the  origi- 
nal triangle.  Therefore,  instead  of  making  the  lines  C  A 
and  C  B  as  long  as  they  were  found  to  be,  we  might  make 
them  one-half  as  long,  or  one-tenth  as  long,  or  we  might 
use  any  convenient  fraction  of  the  actual  length.  In  such 
cases  the  line  found  would  be  the  same  fraction  of  the  orig- 
inal line  A  B.  We  might  go  still  further,  and  use  a  centi- 
meter or  an  inch  to  represent  several  feet,  and  we  then 
could  draw  a  plan  of  the  ground  on  paper.  The  process 
just  described  is  called  "drawing  to  scale."  In  drawing 
to  scale  you  cannot  be  too  careful ;  a  slight  error  makes  a 
great  diiference  ;  for  instance,  if  one  centimeter  stands  for 
ten  feet,  an  error  of  a  millimeter  in  your  plan  would  make 
an  error  of  one  foot  on  the  ground.  In  choosing  your 
scale  it  is  well  to  reduce  the  lines  as  little  as  the  size  of 
your  paper  will  allow. 

How  long  a  line  can  you  represent  on  your  paper,  if  one 
centimeter  represents  ten  feet  ?  How  long  a  line,  if  one 
inch  represents  ten  feet  ? 

184.  Draw  lines  on  your  paper  representing  lengths  of 
150  ft.,  161  ft.,  84  ft.,  26J  ft.,  using  the  scale  1  cm.  =  10 
ft. 

185.  Draw  lines  representing  lengths  of  75  ft.,  63  ft., 
110  ft.,  87  ft.,  using  the  scale  1  in.  =  20  ft. 

186.  Draw  a  triangle,  and  find  the  lengths  of  the  sides 
of  the  triangle  which  it  represents,  if  the  scale  is  1  cm.  = 
10  ft.  ;  if  the  scale  is  1  in.  =  10  ft. 

187.  If  you  are  to  represent  lines  between  300  ft.  and 
400  ft.,  what  scale  will  you  choose  for  your  paper  ? 

188.  When  the  scale  is  1  cm.  =  5  ft.,  what  do  lines 
7.5  cm.,  24  cm.,  63  mm.,  and  2  m.  represent  ? 


38  ELEMENTARY  AND 

189.  If  the  scale  used  is  \  in.  =  3  ft.,  how  long  are  the 
lines  which  are  represented  by  2}  in.,  3|  in.,  7J  in.  ? 

190.  Two  sides,  A  C  and  B  C,  of  a  triangular  field  were 
found  to  be  150.75  ft.  and  200  ft.  respectively,  while  the 
included  angle  was  70°.  Draw  a  plan  of  the  field,  using  the 
scale  1  cm.  =  15  ft.,  and  calculate  the  length  of  A  D. 

191.  Choose  two  points,  A  Mid  B,  in  different,  but  ad- 
joining, school-rooms  in  such  a  way  that  one  point  cannot 
be  seen  from  the  other,  but  that  both  can  be  seen  by  a  per- 
son standing  in  the  doorway  between  the  rooms.  Explain 
what  measurements  you  would  take  to  find  the  distance 
A  B  ;  actually  take  the  measurements,  choose  a  scale  suited 
to  the  size  of  your  paper,  draw  a  plan,  and  find  the  length 
of  the  line  A  B. 

Note. — A  great  many  problems  of  this  kind  can  be 
given  with  profit.  It  is  important,  in  the  first  few  cases  at 
least,  that  the  actual  measurements  be  taken  by  the  pupil. 
Later,  if  it  is  thought  best,  measurements  can  be  taken 
from  city  plans  that  will  enable  the  pupils  to  find  the 
shortest  distance  between  well-known  points.  In  all  cases 
mat  and  accurate  work  should  be  insisted  upon.  It  is  well 
t<>  have  the  pupils  select  scales  adapted  to  the  paper  used, 
and  to  have  the  same  problem  solved  by  different  scales. 

L98.  In  178  we  found  that  two  triangles  would  be  alike 
necessarily,    if    they   were  c 

knOWTI  to  have  two  sides 
and  the  in  eluded  angle 
alike  respectively.  Let  us 
see  if  we  cannot  leave  out 
the  word  included  and 
show  that  two  triangles 
must  be  alike  in  all  re- 
spects, if  they  have  two 
sides  and   an  angle   alike 

respectively.  Make  a  triangle,  ABC,  taking  heed  to  make 
A  C  longer  than  B  C  (Fig.  37).  Make  another  triangle 
R  S  T  havinir— 


CONSTRUCTIONAL  GEOMETRY  39 

1st.  RS^AC; 

2d.  /  S  R  T  =  /  C  A  B  ; 

3d.  S  T  =  C  B. 

After  taking  the  first  two  steps,  are  you  obliged  to  take 
the  third  step,  which  will  complete  your  triangle,  in  one 
way,  or  do  you  have  a  choice  of  ways  ?  Make  the  following 
statement  true  by  crossing  out  the  words  that  need  to  be 
crossed  out :  Two  triangles  are  or  are  not  necessarily  equal, 
if  they  have  two  sides  and  the  angle  opposite  the  smaller 
side  alike  respectively. 

193.  Making  use  of  the  same  triangle  (A  B  C,  Fig.  37), 
make  a  triangle  MNP  having — 

1st.  N  P  =  B  C  ; 

2d./PNM=/CBA; 

3d.  P  M  =  C  A. 

You  remember  that  C  A  was  purposely  made  longer 
than  B  C.  After  taking  the  first  two  steps,  can  you  take 
the  third  step  in  more  than  one  way  ?  What  words  do  you 
think  ought  to  be  crossed  out  in  the  following  statement  : 
Two  triangles  are  or  are  not  necessarily  equal  in  all  re- 
spects, if  they  have  two  sides  and  the  angle  opposite  the 
greater  of  the  two  sides  alike  respectively?  If,  then, 
we  leave  out  the  word  included  in  the  principle  of  178, 
what  must  we  put  in  its  place  ? 

194.  Make  three  different  triangles.  Make  three  new  tri- 
angles by  taking  in  each  triangle  of  the  first  set  two  sides 
and  an  angle  opposite  the  greater  of  the  two  sides  chosen. 

195.  Make  as  many  triangles  as  you  can  by  taking  in 
each  triangle  of  the  first  three  triangles  drawn  in  194  two 
sides  and  the  angle  opposite  the  smaller  of  the  two  sides 
chosen. 

SECTION  IX.     REVIEW. 

196.  What  are  parallel  lines  ? 

197.  If  a  line  on  the  floor  is  parallel  to  a  line  on  the 
ceiling,  could  you  move  one  to  the  position  of  the  other 


40  ELEMENTARY  AND 

without  changing  its  direction  ?    What  would  the  moving 
line  generate  ? 

198.  Can  you  draw  two  parallel  lines  so  that  a  plane 
surface  cannot  be  found  which  will  contain  both  lines  ? 
Upon  what  axiom  does  your  answer  depend  ? 

199.  If  a  line  on  the  floor  is  not  parallel  to  a  line  on  the 
ceiling,  can  you  move  one  line  to  the  position  of  the  other 
without  changing  its  direction  ?  What  would  the  moving 
line  generate  in  reaching  the  position  of  the  second  line  ? 

200.  Draw  several  pairs  of  lines  that  would  not  meet 
within  the  limits  of  your  paper,  if  extended.  Estimate 
the  angles  between  them,  pair  by  pair,  and  record  your  es- 
timate ;  then,  by  the  method  of  168,  draw  a  parallel  to 
one  line  of  each  pair  that  shall  cross  the  second  line,  meas- 
ure the  angle,  and  record  the  true  value  and  the  error.  In 
doing  this  exercise  take  care  to  have  some  of  the  lines 
make  obtuse  angles  and  some  acute  angles. 

201.  What  is  an  oblique  angle  ? 

202.  What  is  a  geometrical  proof  ? 

203.  One  angle  of  a  parallelogram  is  50°  ;  what  are  the 
other  angles  ? 

204.  One  angle  of  a  parallelogram  is  twice  another  ;  what 
are  the  angles  ? 

205.  Extend  side  A  B  of  the  triangle  A  B  C  to  R.  Try 
to  show  that  /CBR  is  equal  to  the  sum  of  two  angles 
of  the  triangle,  /  B  A  C  and  /  A  C  B. 

206.  Repeat  the  problem  of  205,  extending  a  line  at  each 
vertex  of  the  triangle,  and  showing  what  angles  of  the 
triangle  are  needed  to  make  each  exterior  angle  formed. 

207.  One  acute  angle  of  a  right  triangle  is  70°  ;  what  is 
the  other  ? 

208.  Make  a  triangle  ABC,  and  make  another  like  it 
in  as  many  ways  as  you  can,  by  principles  already  estab- 
lished. 

209.  Can  two  triangles  have  two  sides  and  an  angle  re- 
spectively alike  and  still  be  unequal  ? 


CONSTRUCTIONAL  GEOMETRY 


41 


SECTION  X.      ISOSCELES  TRIANGLES. 


210.  Make  a  triangle  ABC  with  A  C  =  B  C.  Such  a 
triangle  is  said  to  have  equal  legs,  and  for  that  reason  is 
called  an  isosceles  triangle,  a 
name  that  in  Greek  means 
"  with  equal  legs."  The  third 
side,  in  this  case  A  B,  is  called 
the  base  of  the  triangle  ;  the 
isosceles  triangle  is  generally 
thought  of  as  standing  on  this 
base,  the  legs  meeting  in  a  point 
which  is  called  the  vertex.  In 
triangles  that  are  not  isosceles 
any  side  may  be  thought  of  as  the  base,  and  no  one  vertex 
is  singled  out  as  the  vertex.  In  an  isosceles  triangle  M  P  N, 
if  P  is  the  vertex,  what  is  the  base,  and  what  are  the  legs  ? 

211.  Triangle  R  S  T  has  R  T=S  T  (Fig.  39).  To  the 
eye  /  R  seems  like  /  S.  Recall  the  tests  for  equal  angles 
thus  far  discovered,  and  decide  whether  any  one  of  them 
will  prove  the  angles  equal.     Do  the  sides  point  the  same 


way  respectively  ?    Do  the  sides  point  in  opposite  direc- 
tions respectively  ?     Are  they  supplements  of   the  same 


42  ELEMENTARY  AND 

angle  ?  Study  the  directions  of  the  sides  of  the  angles 
carefully,  and  be  able  to  give  a  reason  for  your  answer, 
whether  yes  or  no.  If  all  these  tests  fail,  there  is  still  one 
more  chance  for  proving  the  angles  equal,  becausejn  ITU  it 
was  found  that  two  angles  must  be  alike,  if  they  are  cor- 
n-ponding angles  of  triangles  that  can  be  proved  equal. 
At  first  thought  this  test  seems  of  little  use,  because  only 
one  triangle,  K  S  T,  is  mentioned  here  ;  but  the  triangle  can 
be  divided  into  two  triangles  in  a  great  many  ways  by  lines 
running  from  T  to  different  points  in  the  base.  Choose 
from  all  the  possible  lines  the  one  that  divides  the  angle 
B  T  S  in  halves.  You  will  be  obliged  to  use  your  protractor 
to  find  this  line,  although  the  imagination  will  supply  it 
easily.  It  is  represented  by  T  1)  E  in  the  figure.  Two 
triangles,  /\  R  T  E  and  /\  S  T  E,  are  now  formed  ;  to  aid 
the  imagination  it  may  be  well  to  draw  them  apart  from  each 
other  (Fig.  89),  or  to  make  a  paper  isosceles  triangle  and 
to  cut  it  along  the  line  corresponding  to  T  D  R, 

In  comparing  ^  R  T  E  and  E  T  8  we  know — 

1st,  that  T  R  =  T  S  from  the  nature  of  isosceles  trian- 
gles; 

\M.  that  T  B    -  'I'  B  from  the  nature  of  straight  lines  ; 

3d,  that  /  R  T  E  =  /  E  T  S  because  each  is  one-half  of 
B  T  S  :  so  that  our  knowledge  of  these  two  triangles  is  based 
upon  the  nature  of  the  figures  and  not  upon  our  measure- 
ments. 

Are  the  three  things  that  we  know  about  the  triangles 
enough  to  prove  them  equal  ?  Upon  what  principle  of  equal 
triangles  do  you  rely  ?  Write  out  the  remaining  parts 
of  the  two  triangles  that  must  be  equal  respectively  from 
the  principle  stated  in  179.  You  know  that  there  must  be 
in  /\  E  T  S  an  angle  equal  to  /  R  of  /\  E  T  K,  because 
the  two  triangles  are  alike  ;  but  how  do  3*011  know  whether 
/  S  or  /  E  is  the  one  equal  to  /  R  ?  In  this  particular 
case  there  can  be  no  doubt,  because  /  E  evidently  is 
much  larger  than  /  R  ;  but  it  is  not  safe  to  rely  upon  the 
eye  in  such   matters.     A  method  of  selecting  the  corre- 


CONSTRUCTIONAL  GEOMETRY  43 

sponding  angles  and  lines  that  can  never  fail  is  based  upon 
the  fact  that  the  corresponding  angles  are  opposite  the 
sides  that  have  been  proved  equal,  and  the  correspond- 
ing sides  are  opposite  the  angles  that  have  been  proved 
equal,  fn  this  case  E  T  has  been  proved  equal  to  E  T  : 
opposite  ETin/\ETRis/R;  opposite  E  T  in  /\  E  T  S 
is  /  S  ;  /  R  and  /  S,  therefore,  are  corresponding  angles 
of  the  two  triangles,  and  are  equal  by  179.  Select  the  other 
corresponding  parts  by  the  same  kind  of  test.  You  cannot 
acquire  too  much  skill  in  selecting  corresponding  parts  of 
figures. 

212.  The  knowledge  that  we  have  gained  in  211  about 
isosceles  triangles  gives  us  a  fifth  test  for  detecting  equal 
angles  without  protractor  or  compasses.  We  now  know 
that  the  base  angles  of  an  isosceles  triangle  must  be 
equal. 

213.  In  showing  that  /  R  =  /  S  (Fig.  39)  we  discovered 
a  fact  that  is  of  even  greater  importance,  namely,  that 
/RET=/TES.  What  kind  of  angle  must  each 
one  be  ?  Why  ?  State  the  fact  as  a  principle  relating  to 
the  bisector  of  the  vertical  angle  of  an  isosceles  triangle.  A 
principle  that  is  discovered  while  we  are  trying  to  prove 
another  principle  is  called  a  corollary. 

214.  There  is  a  second  corollary  resulting  from  the  proof 
in  211  about  the  point  E,  in  which  the  bisector  of  the  verti- 
cal angle  strikes  the  base.  State  it,  and  record  the  state- 
ment decided  by  the  class  to  be  the  best. 

215.  Can  you  without  your  protractor  tell  the  number  of 
degrees  in  any  angle  of  /\  E  T  R  ? 

216.  If  /  R  of  A  R  T  S  (Fig.  39)  is  70°,  what  are  the 
angles  at  S  and  T  ?     Why  ? 

217.  What  are  the  base  angles  of  an  isosceles  triangle,  if 
the  vertical  angle  is  70°  ?  if  it  is  60°  ?  if  it  is  90°  ? 

218.  If  the  base  angle  of  an  isosceles  triangle  is  twice  the 
vertical  angle,  what  is  the  value  of  each  angle  of  the  tri- 
angle ? 

219.  Make  a  triangle  with  all  three  sides,  or  legs,  equal. 


44 


ELEMENTARY  AND 


1*0' 


-X. 


B 


Ki.i  ML 


What  can  yon  say  about  its  angles  ?     Why  ?    Sncli  a  tri- 
angle is  called  an  equilateral  triangle.     Why  ? 

220.  The  vertical  angle  of  an  isosceles  triangle  is  twice 
the  base  angle  ;  can  you  tell  the  degrees  in  each  angle  ? 

221.  In  Fig.  40,  ^  ABC  is 
an  isosceles  triangle ;  /CBD 
formed  by  extending  the  base 
A  B  is  140°.  What  are  the 
angles  of  /\  A  B  C  ? 

222.  In  Fig.  41/\MNS 
is  an  isosceles  triangle  whose 
base  angle,  M,  is  60°.     M  S  is  extended  so  that  S  P  =  MS. 
Can    yon  name    in   degrees  all  the 
angles    of    the    figure  P       What    is 
/  PM  Iff 

223.  What  would  be  the  value  of 

1  V  N  M  (Fig.  41)  if  we  should  take 

2  M  =  50°  ?  What,  if  /  II  =  70°  ? 
if  /  M  =  .v.  Can  you  discover  any 
principle  about  /  V  S  M  ': 

224.  When  two  lines  form  a  right 
angle  thej  are  said  to  be  perpendicu- 
lar to  each  other.  The  sign  for  per- 
pendicular is  J_. 

What  examples  of  perpendicular 
lines  have  we  had  in  Section  X.  ? 
(;i\c  examples  of  perpendicular  lines 
in  the  room  about  you. 

225.  Can  you  use  the  principle  of  223  to  draw  a  perpen- 
dicular to  a  line  A  B  at  B  ?  Suggestion  :  Choose  a  point 
C  corresponding  to  S  in  Fig.  41,  and  outline  the  necessary 
isosceles  triangles. 

'.'-.  Draw  six  different  lines,  and  draw  perpendiculars  to 
them,  using  the  method  of  225.  Reduce  the  construction 
lines  as  much  as  you  can  in  the  last  two  or  three  cases.  If 
you  thoroughly  understand  the  principle,  you  can  so  reduce 
the  help  lines  that  they  will  hardly  be  noticed. 


Flo.  4L 


CONSTRUCTIONAL  GEOMETRY 


45 


227.  In  proving  the  fact  about  the  base  angles  of  an 
isosceles  triangle  (211),  it  was  necessary  for  you  to  imag- 
ine the  vertical  angle  bisected,  or  else  to  use  the  protrac- 
tor to  bisect  it.  For  our  proof  the  actual  bisecting  line 
was  not  needed  (why  ?)  ;  but,  as  a  matter  of  practical  con- 
struction, it  is  important  to  be  able  to  bisect  an  angle, 
and  a  construction  problem  is  not  looked  upon  as  solved 
in  Geometry,  unless  the  ruler  and  compasses  are  the  only 
tools  used.  Can  you  bisect  an  angle  with  ruler  and  com- 
passes only  ?  Can  you  prove  that  your  method  is  a  sound 
one  ? 

If  unable  to  bisect  an  angle  for  yourselves,  study  the  fig- 
ure given  here  (Fig.  42),  and  try  to  give  reasons  for  the 
statements  made. 

1st.  With  your  compasses  make  C  A  =  C  B,  using  C  as 
centre. 

2d.  With  A  and  B  as 
centres,  make  A  K  =  B  K, 
putting  K  below  A  B. 

3d.  Join  C  K  and  A  B, 
crossing  each  other,  at  D. 

l./CAB=/CBA. 
Why? 

2.  /DAK=  £  DBK. 
Why? 

3.  £  CAK=  /CBK. 
Why? 

4.  ACAK  =  ACBK. 
Why? 

5./DCA=/DCB. 
Why? 

If  the  five  statements  made  can  be  shown  to  be  necessar- 
ily true,  there  can  be  no  doubt  that  the  line  C  D  K  bisects 
the  angle  A  C  B. 

228.  Make  six  different  angles,  and  bisect  them.  In  the 
first  four  cases  go  through  the  steps  carefully,  giving  all 
the   reasons   which   prove   in   the  end   that  the   angle   is 


46 


ELEMENTARY  AND 


bisected.     In  the  last  two  cases  reduce  the  help  lines  to 
those  which  arc  absolutely  necessary. 

229.  Draw  a  triangle  and  bisect  each  angle. 

230.  Draw  a  triangle,  ABC.     Extend  A  B,  B  C,  and 
C  A.     Bisect  the  external  angles  formed. 

231.  Could  K  have  been  between  C  and  the  line  A  B  in 
227  ?  Study  Fig.  43,  where  C  A  =  C  B,  and  D  A  ==  D  B, 
before  answering.  Are  enough 
facts  known  about  the  figure  to 
prove  that  A  0  A  D  =  AC  B  D? 
If  so,  what  follows  about  an- 
gles A  C  D  and  B  C  D  ? 

232.  Now  that  we  know  how 
to  bisect  an  angle  without  a  pro- 
tractor, we  know  also  how  to 
draw  a  perpendicular  to  a  line 
(see  213)  in  a  new  way,  ami 
how  to  bisect  a  line  at  the  same 
time  (see  214).  Give  an  illustra- 
tion of  bisecting  a  line  by  a  per- 
pendicular line,  explaining  I 
step  taken. 

>.   Make  six  different  lines 
bisect  them   with 
method  described 


Wm  m 


in  various  positions  ami 
perpendicular  lines  according  to  the 
in  232.  Leave  to  the  imagination  all 
lines  and  points  not  absolutely  necessary  for  your  work,  but 
be  able  to  tell  what  lines  need  to  be  imagined. 

234.  Make  a  triangle  and  find  the  middle  point  of  each 
side.  Do  you  notice  anything  ]>eculiar  about  the  bisecting 
perpendiculars  when  they  are  prolonged  ?  Experiment 
with  four  different  triangles,  making  at  least  one  of  the 
triangles  an  obtuse-angled  triangle,  and  note  whether  or  not 
there  is  the  same  peculiarity  always. 

235.  Make  an  equilateral  triangle,  and  draw  semi-circum- 
ferences without  the  triangle  on  each  side  as  a  diameter. 

236.  Draw  a  circumference  about  each  side  of  a  square, 
A  B  C  D,  as  a  diameter. 


CONSTRUCTIONAL  GEOMETRY  47 

237.  Can  you,  by  the  method  of  232,  draw  a  perpendicu- 
lar to  the  line  A  B  at  any  point  of  A  B  except  the  midde 
point  ?     Give  three  illustrations. 

238.  Draw  a  line,  A  B,  choose  a  point  P  (not  the  centre) 
on  A  B,  and  draw  a  perpendicular  to  A  B  at  P,  first  by 
the  method  used  in  226,  and  then  by  the  method  just  ex- 
plained. 

239.  Can  you  use  the  method  of  232  to  draw  a  perpen- 
dicular to  A  B  from  a  point  P  outside  of  A  B  ?  Give  three 
illustrations. 

240.  Draw  a  line,  A  B,  and  choose  a  point,  P,  above  and 
beyond  the  end  of  A  B.  How  will  you  proceed  to  draw  a 
perpendicular  to  A  B  from  P  in  this  case  ? 

241.  Draw  a  triangle,  ABC.  From  each  vertex  draw  a 
perpendicular  to  the  opposite  side.  Do  you  notice  any 
peculiarity  of  the  three  perpendiculars  ?  Repeat  the  ex- 
periment three  or  four  times,  taking  an  obtuse-angled 
triangle  twice. 

242.  The  lines  drawn  in  241  from  the  vertices  of  a  tri- 
angle perpendicular  to  the  opposite  sides  are  called  the 
altitudes  of  the  triangle,  because  they  represent  the  heights 
of  the  triangle  as  it  stands  on  its  different  bases.  Does  the 
altitude  of  a  triangle  ever  join  the  vertex  with  the  middle 
of  the  base  ?  Is  the  altitude  ever  as  long  as  a  side  of  the 
triangle  ? 

243.  With  the  aid  of  the  isosceles  triangle,  wholly  or 
partly  drawn,  lines  and  angles  can  be  bisected,  perpendicu- 
lars to  lines  can  be  drawn,  and  equal  angles  may  be  dis- 
covered. You  cannot,  therefore,  become  too  familiar  with 
what  has  been  shown  about  the  isosceles  triangle.  Try  to 
answer  the  following  test  questions  without  reference  to 
your  note-books.  If  you  are  obliged  to  look  up  the  answers, 
you  will  know  that  you  have  not  sufficiently  mastered  the 
subject,  and  that  you  have  not  been  studying  in  the  right 
way. 

1st.  If  two  isosceles  triangles  have  the  same  base,  what 
can  you  say  about  the  line  that  joins  their  vertices  ? 


48  ELEMENTARY  AND 

2d.  How  many  truths  can  you  give  about  the  line  that 
bisects  the  vertical  angle  of  an  isosceles  triangl. 

3d.  If  you  should  draw  three  different  isosceles  trial Lgi« 
on  the  same  base,  could  you  say  anything  about  the  positions 
of  their  vertices  ? 

4th.  If  a  base  angle  of  an  isosceles  triangle  is  20°,  what 
are  the  remaining  angles  ?  If  the  vertical  angle  is  20°, 
what  are  the  remaining  angles  ? 

■'II.  Make  a  straight  angle.  How  does  this  differ  from 
making  a  straight  line  ?  Bisect  and  also  quarter  the 
straight  angle. 

845.  Make  an  angle  of  G0°  without  your  protractor; 
then  divide  it  into  four  equal  angles.  Divide  a  semi-cir- 
cumference into  twelve  equal  parts. 

2  1  ;.  Divide  a  line,  A  B,  into  four  equal  parts  ;  into  eight 
equal  parts.  Can  you  divide  it  into  three  equal  parts  with 
tin-  aid  of  isosceles  triangl  i 


SECTION  XL     TRIANGLES— Coxti  mi  i». 

c  p 


247.  Make  a  triangle,  ABC  (Fig.  44).     Make  a  second 
triangle,  M  N  P,  as  follows : 
1st.  Make  M  N  =  A  B. 
2d.  Make   /  M  =  /  A. 
3d.  Make  /  N  =  /  B,  thus  completing  A  M  X  P. 


CONSTRUCTIONAL   GEOMETRY  49 

Over  what  parts  of  the  triangle  M  N  P  have  you  had 
control,  and  over  what  parts  have  you  had  no  control  ? 

Can  you  think  of  any  reason  why  /  P  must  be  equal  to 
/  C  ?  Review  178,  and  then  show  that  /\MP,  if 
imagined  to  be  placed  upon  /\ABC,  would  exactly  coin- 
cide with  /\ABC.  Describe  with  great  care  the  placing 
of  /\  M  N  P,  and  show,  by  giving  reasons,  what  must  be  the 
position  of  the  rest  of  the  triangle  after  you  have  placed 
two  of  its  points. 

248.  Can  you  complete  /\  M  N  P  by  making, 
1st.   M  N  =  A  B  ; 

2d.    I  M  =  I  A  ; 

3d.   I  P  =  /  C  ? 

You  will  at  first  be  puzzled  to  decide  where  to  put  the 
vertex  of  the  angle  like  /  C ;  but  a  little  ingenuity  will 
enable  you  to  find  the  right  position  without  measuring 
A  C,  B  C,  or  /  B.     See  155,  if  you  need  a  hint. 

Will  this  triangle  necessarily  be  equal  to  the  triangle 
ABC?  Show,  by  placing  /\MNP  upon  A  A  B  C,  wheth- 
er or  not  it  will  exactly  coincide  with  it. 

249.  As  a  result  of  247  and  248  make  the  following 
statement  true,  and  record  the  corrected  statement  in  the 
summary  : 

Two  triangles  are  or  are  not  necessarily  equal  in  all  re- 
spects, if  two  angles  and  a  side  of  one  are  equal  respectively  to 
two  angles  and  a  corresponding  side  of  the  other. 

250.  Can  you  construct  a  triangle  that  shall  have  a  side 
A  B,  and  its  angles  equal  to  those  of  /\  A  B  C,  and  yet  be 
unequal  to  A  A  B  C,  because  the  side  equal  to  A  B  is  not  in 
a  corresponding  position  ?  How  many  such  triangles 
can  you  construct  ? 

251.  Make  any  two  angles  whose  sum  will  not  amount 
to  a  straight  angle  ;  call  these  angles  x  and  y ;  also  make 
any  line  a.  Construct  as  many  triangles  as  you  can  that 
shall  have  the  angles  x  and  y  and  the  line  a.  Can  you 
make  two  unequal  triangles  that  shall  have  the  side  a  be- 
tween the  angles  x  and  y  ? 

4 


50  ELEMENTARY  AM> 

252.  A  person  wishing  to  find  the  distance  from  his 
house  on  the  sea-shore  to  an  island,  sighted  from  his 
house,  represented  by  point  A  in  Fig.  45,  a  prominent  ob- 
ject, B,  on  the  island,  and  also  a  place,  C,  on  land,  which 


could  be  easily  reached,  and  from  which  B  could  be  seen  ; 
after  noting  the  angle  BAG,  he  measured  A  C,  and 
sighted  B  and  A  from  C,  noting  the  angle  A  C  B.  With 
these  measurements  before  him,  he  constructed  on  paper 
the  triangle  MNP,  having  /  M  =  /  A,  and  /  N  =  /  C. 
How  did  he  estimate  the  distance  A  B  ? 

253.  Let  the  class  calculate  the  distance  from  the  school- 
house  to  a  tree  which  can  be  seen  from  two  windows  of  the 
room,  making  the  proper  measurements  with  as  few  sug- 
gestions as  possible  from  the  teacher.  The  measurements 
may  be  taken  in  the  class,  and  the  calculations  may  be 
made  later,  if  the  time  seems  too  short. 

254.  What  additional  measurement  would  be  needed  to 
determine  the  height  of  the  tree  ? 

255.  Describe  and  take  the  measurements  that  would  be 
necessary  to  determine  the  length  of  a  ladder  that  would 
reach  to  a  particular  branch  on  the  same  tree,  if  the  ladder 
rests  on  the  ground  five  feet  from  the  tree. 

256.  A  ship  captain  sights  a  light-house  at  an  angle  of 
70°  from  his  course.  After  sailing  three  miles  on  the 
same  course,  he  sights  it  airain  at  an  angle  of  100°  from 
his  course.     How  far  was  he  from  the  light-house  each 


CONSTRUCTIONAL  GEOMETRY  51 

time  ?     How  far,   when  he  passed  nearest  to  the  light- 
house ? 

257.  Observe  two  points  on  a  house  or  tree,  one  of  which 
is  directly  over  the  other.  Take  the  measurements  that 
will  enable  you  to  calculate  the  distance  between  the  two 
points  without  leaving  the  room. 

258.  Can  you  think  of  any  reason  why  you  can  not  use 
the  method  used  in  256  to  find  the  distance  to  a  star  ? 

259.  A  man,  standing  on  the  bank  of  a  stream  at  a  point 
A,  sees  at  an  elevation  of  60°  the  top  of  a  tree,  which  stands 
at  a  point  B  on  the  opposite  shore  ;  he  walks  away  from  the 
shore  on  the  line  B  A  for  80  ft.  ;  he  then  sees  the  top  of 
the  same  tree  at  an  elevation  of  30°.  Draw  a  plan  to  scale, 
and  find  the  height  of  the  tree  and  the  width  of  the  river. 

260.  Construct  a  right  triangle  with  one  leg  3  in.  and 
the  adjacent  acute  angle  50°. 

261.  Construct  a  triangle  having  one  angle  30°,  another 
65°,  and  the  side  joining  the  vertices  of  these  angles  3J  in. 

262.  Construct  a  triangle,  ABC,  with  /  A  =  70°,  /  B 
=  60°,  and  side  B  C  =  2  in. 

263.  How  would  you  find  the  distance  between  two 
islands  without  leaving  the  main  land  ? 

264/  In  measuring  a  triangular  field,  A  B  C,  a  man  found 
the  side  A  B  to  be  300  yds.  ;  he  found  the  angle  A  B  C  to 
be  95° ;  when  he  reached  C,  he  was  not  satisfied  with  his 
measurement  of  B  C.  Instead  of  retracing  his  steps  to  meas- 
ure B  C,  he  sighted  A  from  C,  finding  the  angle  A  C  B  to  be 
40°.  How  did  this  help  him 
to  find  B  C  ?    What  was  the  £ 

length  of  B  C  ?  of  A  C  ?  /\ \ 

265.    Construct  a  triangle,  yS  N. 

A  B  C,  by  making  A  B  any     yS  ^^ 

desired  length,   and  by  mak-   £ jjj ^ 

ing    /  B  =  /  A.      Will   this  ^ 

triangle  have  equal  legs,  A  C 

and  B  C  ?    If  you  determine  your  answer  by  the  appear- 
ance of  the  figure,  or  by  testing  the  lengths  with  your  com- 


52  ELEMENTARY  AND 

passes,  you  will  doubtless  answer  that  A  C  and  B  C  are 
equal.  But  is  there  any  reason  in  the  nature  of  things  why 
they  must  be  equal  ?  Let  us  investigate.  Imagine  /  A  C  B 
bisected  by  a  line,  C  D,  and,  to  aid  the  imagination,  draw 
the  bisector  C  D  by  the  method  of  227.  Examine  carefully 
^ADC  and  BDC.  Point  out  what  parts  of  the  first 
have  their  equals  in  the  second.  Are  there  enough  equal 
parts  to  make  the  triangles  necessarily  equal  in  all  respects  ? 
What  principle  of  equal  triangles  do  you  rely  upon  ?  After 
satisfying  yourselves  that  the  triangles  are  equal,  arrange 
your  work  as  follows  : 

In  A  A  D  0  and  A  B  D  c  &  is  known, 

1st,  that  /  A  =  /  B  (by  construction) ; 

2d,  that  /ACD=/BCD  (each  /  =  J  of  /  A  C  B) ; 

3d,  that  C  D  =  C  D  (same  line). 

.-.  AABC  =  ABCD. 

(Quote  principle  that  you  rely  upon  here.) 

.*.  A  C  =  B  C  (give  reason  here). 

.  •.  A  A  C  B  is  isosceles. 

.'.  A  triangle  must  be  isosceles,  if  its  base  angles  are 
equal. 

266.  Why  could  we  not  have  said  at  once  in  265  that  AD 
=  B  D,  and  that  /CDA=/CDB,  thus  enabling  ofl  to 
use  a  different  principle  for  testing  the  equality  of  the  tri- 
angles ? 

267.  Make  a  triangle  with  its  base  angles  60°.  What 
must  the  third  angle  be  ?     What  kind  of  triangle  is  it  ? 

268.  From  a  ship  that  was  sailing  in  a  straight  course, 
A  B,  a  light  was  sighted  at  an  angle  of  40°  with  the  course 
A  B  ;  after  the  ship  had  gone  four  miles,  the  same  light 
was  sighted  at  an  angle  of  70°  with  B  A.  How  far  was  the 
ship  from  the  light  at  the  first  observation  ? 

269.  Make  a  right  triangle,  ABC,  with  right  angle  at 
C  and  with  acute  angle  A  equal  to  60°.  How  many  degrees 
in  angle  B  ?  Draw  a  line,  C  K,  in  such  a  way  that  /  K  C  B 
=  /  K  B  C.  Study  the  figure  thus  formed,  and  select  all 
the  angles  whose  value  in  degrees  you  know  without  the  aid 


CONSTRUCTIONAL  GEOMETRY  53 

of  a  protractor  ;  also  select  any  lines  that  you  can  show  to 
be  alike  without  measurement.  Do  you  discover  any  truth 
relating  to  the  longest  and  shortest  lines  of  a  right  triangle 
one  of  whose  angles  is  60°  ?  Note  the  distance  of  K  from 
A,  B,  and  C.  Compare  what  you  note  in  this  case  with 
what  you  learned  in  225.     K  is  on  A  B. 

270.  Repeat  the  experiment  made  in  269,  this  time 
making  /  A  =  70°.  Make  /KCB^/KBCas  before. 
Give  the  value  of  all  the  angles  formed.  Can  you  say  the 
same  thing  about  K  that  you  did  in  269  ? 

271.  Make  a  right  triangle,  ABC,  without  knowing  the 
exact  value  of  acute  angle  A.  Draw  C  K  as  before  so  that 
/KCB=  /KBC.  Calling  /_  A  =  /  x,  can  you  name 
all  the  angles  formed  by  means  of  /  x  9  Can  you  show 
that  what  you  learned  about  K  in  269  and  270  still  holds 
true  ? 

272.  If  with  the  middle  point  of  the  longest  side  or 
hypotenuse  of  a  right  triangle  as  a  centre,  you  describe  a 
circumference,  will  it  pass  through  the  vertices  of  the 
right  triangle  ?  Why  ?  The  principle  involved  in  this 
problem  plays  an  important  part  in  later  problems,  so  that 
it  is  well  to  record  it,  and  to  make  special  effort  to  remem- 
ber it.  It  may  be  stated  thus  :  The  middle  point  of  the 
hypotenuse  of  a  right  triangle  is  the  same  distance  from  the 
three  vertices  of  the  right  triangle. 

273.  If,  when  walking  on  a  straight  level  road,  A  B,  you 
sight  from  A  a  tree  across  a  field  at  an  angle  of  70°  with 
A  B,  at  what  angle  from  B  A  will  you  have  to  sight  the 
tree  to  know  that  you  have  walked  a  distance  equal  to  that 
from  A  to  the  tree  ? 

274.  Wishing  to  cut  down  a  tall  tree,  a  man  found  that 
it  was  of  great  importance  to  know  how  far  the  tree  would 
reach  when  felled.  Having  no  instruments  with  him,  he 
fastened  two  straight  sticks  together  at  an  angle  of  45°. 
How  could  he  get  an  angle  of  45°  degrees  with  only  a  piece 
of  string  and  a  stake  to  work  with  ?  Can  you  describe 
how  he  used  the  simple  instrument  to  find  the  height  of 


54  ELEMENTARY  AND 

the  tree  and  the  distance  wanted  ?    Why  did  he  choose 
the  angle  45°  ? 

275.  How  many  ways  of  recognizing  equal  angles  without 
measurement  have  you  learned  thus  far  ?     Give  them  all. 

276.  How  many  ways  of  recognizing  equal  lines  without 
measurement  do  you  know  ? 

277.  Give  the  three   tests    for    recognizing   equal  tri- 
angles. 

278.  Make  a  triangle  (A  B  C,  Fig.  47).     Make  M  N  = 


B       M 


Fio.  47. 


A  B.  With  M  as  centre  and  with  A  C  as  radius  describe 
an  arc.  With  N  as  centre  and  B  C  as  radius  describe  an 
arc  cutting  the  first  arc  at  P  and  Q.  Complete  the  tri- 
angles M  P  1ST  and  MQN.  Clearly  these  triangles  have 
the  same  three  sides  that  triangle  ABC  has.  Is  there 
any  reason  for  thinking  that  /\  M  N  P  has  the  same  angles 
as/\MNQ?  If  only  one  angle  is  the  same  in  both,  the  tri- 
angles must  be  equal  in  all  respects.  Why  ?  Let  us  aim 
to  show  that  /MPN=/MQN.  Join  P  Q.  What 
kind  of  triangle  is^PMQ?    Why  ? 

.-.   /  MPQ=  /_ Why? 

Likewise  /NPQ=/ Why  ? 

.-.   /MPN=/ Why? 

.-.   AMpN=AMQN.    Why? 
Should  you  place  /\  A  B  C  in  the  plane  of  the  paper  with 


CONSTRUCTIONAL   GEOMETRY  55 

A  B  upon  its  equal  M  N,  where  would  C  have  to  fall  ? 
Why?  Hence/\ABC  =  AMNP  =  AMQN-  Hence 
a  new  principle  for  recognizing  equal  triangles  :  Two  tri- 
angles are  equal  in  all  respects,  if  the  three  sides  of  one  are 
equal  to  the  three  sides  of  the  other  respectively. 

279.  In  Fig.  47  can  you  tell  the  number  of  degrees  in 
the  angle  between  M  N  and  PQ?     Give  reasons. 

280.  If,  in  A  A  B  C,  A  B  is  7  in-  lon&  and  A  C  is  5  in. 
long,  how  short  can  B  C  be  ?  How  long  can  it  be  ?  What 
reasons  can  you  give  for  your  answers  ? 


SECTION  XII.     QUADRILATERALS. 

281.  Construct  quadrilaterals  with  your  compasses  and 
ruler  as  follows : 

1st.  A  quadrilateral  no  two  of  whose  sides  are  parallel. 
*.  Such  a  quadrilateral  is  called  a  trapezium. 

2d.  A  quadrilateral  only  two  of  whose  sides  are  parallel. 
This  figure  is  a  trapezoid. 

3d.  A  quadrilateral  whose  opposite  sides  are  parallel. 
What  is  this  figure  called  ? 

282.  If  one  line  crosses  two  others,  what  angles  must  be 
alike  to  make  the  lines  parallel  ?  Is  there  more  than  one 
pair  of  angles  whose  equality  will  make  the  lines  parallel  ? 

283.  How  many  degrees  must  there  be  in  the  sum  of  the 
four  angles  of  a  quadrilateral  ?     See  106-110. 

284.  Can  you  make  a  trapezium  with  two  of  its  sides 
alike  ?  with  three  of  its  sides  alike  ?  with  all  four  of 
its  sides  alike  ?  In  the  last  case,  draw  a  line  joining  two 
opposite  corners,  dividing  the  quadrilateral  into  two  tri- 
angles, and,  from  your  study  of  these  triangles,  decide 
whether  the  figure  may  be  a  trapezium  or  must  be  another 
kind  of  quadrilateral. 

285.  Can  you  make  a  trapezium  with  two  angles  alike  ? 


56  ELEMENTARY  AND 

with  three  angles  alike  ?  with  four  angles  alike  ?  Can 
you  make  a  trapezium  with  three  of  its  angles  60°  each  ? 
Why  ?     Can  there  be  three  angles  of  70°  in  a  trapezium  ? 

286.  Can  you  make  a  trapezoid  with  two  sides  alike  ? 
with  three  sides  alike  ?  with  two  angles  alike  ?  with  three 
angles  alike  ? 

287.  Draw  a  trapezoid,  A  B  C  D,  making  A  B  and  C  D 
antiparallel.  It  is  possible  to  fix  the  value  of  /  D  from 
the  value  of  /  A.  Do  you  see  how  ?  If  /  A  =  70°, 
what  is  /  D  ?  If  /  B  =  25°,  what  is  /  C  ?  Does  /  A 
help  you  to  find  /  C  ? 

.  288.  If,  in  making  a  trapezoid  A  B  C  D,  you  first  make 
/  A  =  70°  and  /  B  =  40°,  can  you  make  fe  C  and  D 
of  any  size  that  you  please,  or  are  they  fixed  in  value  ? 
Give  a  reason  for  your  answer. 

289.  How  many  angles  of  a  parallelogram  can  you  make 
before  all  of  the  angles  are  fixed  in  value  ? 

290.  The  most  interesting  of  the  quadrilaterals  are  the 
parallelograms,  of  which  there  are  several  varieties.  Those 
with  their  sides  all  alike  and  with  their  angles  90°  each  are 
called  squares. 

Those  that  have  their  sides  alike  but  their  angles  ob- 
lique are  called  rhombuses.  Those  that  have  right  angles 
but  adjacent  sides  unequal  are  called  rectangles.  Those 
that  have  oblique  angles  and  adjacent  sides  unequal  are 
called  rhomboids. 

291.  Construct  models  of  each  kind  of  parallelogram, 
describing  the  steps  in  the  order  in  which  you  take  them 
and  at  the  time  at  which  you  take  them. 

292.  Select  examples  of  the  different  kinds  of  parallelo- 
gram from  the  school-room.  What  kind  of  parallelogram 
is  a  sheet  of  writing-paper  ?  a  diamond  ?  a  window-pane  ? 

293.  Can  you  make  a  rhomboid  with  three  of  its  angles 
alike  ?    Why  ? 

294.  Is  any  proof  needed  to  show  that  the  opposite  sides 
of  a  parallelogram  are  parallel,  pair  by  pair  ?     Why  ? 

295.  Can  you  draw  a  parallelogram  whose  opposite  sides 


CONSTRUCTIONAL  GEOMETRY  57 

are  not  equal  ?      Does  your  answer  to  this  need  proof  ? 
Why? 

296.  Can  you  see  any  reason  why  a  parallelogram  should 
not  be  defined  in  the  first  place  as  a  quadrilateral  with  its 
opposite  sides  parallel  and  equal  f 

Note. — It  is  important  at  this  point  to  make  clear,  if 
possible,  the  danger  of  imposing  too  many  conditions  upon 
lines  and  angles,  and  also,  in  the  same  line  of  thought,  to 
impress  upon  the  pupil  the  importance  of  drawing  the 
figures  exactly  according  to  the  conditions,  except  in  rare 
cases.  If  the  condition  imposed  is  that  the  lines  be  made 
parallel,  they  should  be  made  parallel,  and  not  equal,  al- 
though it  may  be  possible  to  prove  that  they  are  parallel 
when  made  equal.  The  subject  should  not  be  left  until, 
by  varied  illustration,  some  impression  is  produced  upon 
the  pupil.  It  will  generally  be  found  necessary  to  return 
to  the  subject  at  short  intervals.  Abundant  opportunity 
for  testing  the  pupil's  understanding  of  the  subject  may 
be  found  in  questions  about  the  line  bisecting  the  vertical 
angle  of  an  isosceles  triangle. 

297.  If  a  parallelogram  really  cannot  be  drawn  with  its 
opposite  sides  unequal,  there  must  be  some  reason  for  the 
fact  based  upon  the  nature  of  lines  and  angles,  or  upon 
principles  already  established.  Draw  an  angle,  DAB; 
complete  a  parallelogram  by  making  D  C  f  f  A  B  and 
B  G  f  f  A  J),  taking  pains  to  make  A  B  and  A  D  un- 
equal. Draw  a  line,  D  B,  joining  two  opposite  vertices 
(the  line  is  called  a  diagonal),  and  study  the  triangles  thus 
formed.  It  should  be  a  simple  matter  for  you  to  prove 
/\ABD=/\BCD.  Arrange  your  work  in  a  vertical 
line,  putting  no  more  than  one  fact  and  its  reason  on  a 
line.     Follow  this  model : 

I.  I  make  D  C  f  f  A  B. 

II.  I  make  B  C  f  f  AD,  completing  a  parallelogram. 
To  prove  A  B  =  D  C  and  A  D  =  B  C. 
III..  Draw  diagonal  D  B  and  compare  ^A  B  D  and  D  B  C. 
IV.    / =  / (Insert  reason.) 


58  ELEMENTARY  AND 

V.    /   =  / (Reason.) 

VI.  Side =  Side (Reason.) 

VII.  .-.  A  ••••  =  A  ••••  (Reason.) 
VIII.  .\,  etc. 

298.  It  may  help  you  to  understand  297  to  cut  a  paral- 
lelogram something  like  A  B  C  D  from  stiff  paper,  and  to 
cut  the  parallelogram  into  two  parts  along  the  diagonal 
D  B.  Can  you  place  the  two  triangles  formed  by  cutting 
the  paper  parallelogram  so  that  they  will  not  form  a  paral- 
lelogram ?  Is  a  quadrilateral  whose  diagonal  divides  it 
into  two  equal  triangles  necessarily  a  parallelogram  ? 

299.  Can  you  place  the  two  paper  triangles  of  298  in 
more  than  one  position  so  that  they  will  form  a  parallelo- 
gram ? 

300.  Does  it  make  any  difference  which  diagonal  you 
draw  ?  Will  the  parallelogram  be  divided  into  two  equal 
triangles  in  each  case  ? 

301.  How  does  a  rhomboid  differ  from  a  rhombus  ? 

302.  How  does  a  rhombus  differ  from  a  square  ? 

303.  How  does  a  rectangle  differ  from  a  square  ? 

304.  Notice  that,  although  squares,  rhombuses,  and  rec- 
tangles are  not  rhomboids,  they  differ  from  the  rhomboid 
merely  in  having  special  characteristics  in  addition  to 
those  of  the  rhomboid.  Hence  any  general  principle  that 
is  proved  about  the  rhomboid  is  applicable  to  the  rhombus, 
the  square,  and  the  rectangle;  but  it  is  dangerous  to  as- 
sume that  a  principle  proved  true  of  a  square,  a  rhombus, 
or  a  rectangle  can  be  applied  to  a  rhomboid,  because  the 
principle  may  depend  upon  the  special  characteristics  of  the 
former  figures.  Therefore,  when  trying  to  prove  a  prin- 
ciple that  will  apply  to  all  .parallelograms,  be  careful  to 
draw  a  rhomboid  for  illustration,  or  to  have  a  rhomboid  in 
your  imagination. 

305.  Into  what  kind  of  triangles  does  a  diagonal  divide  a 
square  ?  a  rectangle  ?  a  rhombus  ? 

306.  Can  you  construct  a  parallelogram  with  one  angle 
90°  and  another  80°  ?    Why  ? 


CONSTRUCTIONAL   GEOMETRY  59 

307.  If  the  adjacent  sides  of  a  parallelogram  are  equal, 
is  there  any  doubt  as  to  what  kind  of  parallelogram,  the 
figure  is  ? 

308.  Can  you  make  a  rhomboid  with  one  side  3  in.,  an- 
other 2  in.,  and  a  diagonal  5J  in.  ?     Why  ? 

309.  Make  a  rhomboid,  A  B  0  D  (Fig.  48).  What  pre- 
cautions must  you  take  to  be  sure  that  the  figure  is  a 
rhomboid  ?      Draw  both 

diagonals,  A  C  and  B  D, 
crossing  at  M. 

1st.  Make  a  record  in  a 
column  as  in  297  of  all 
the  pairs  of  equal  angles 
that  you  can  find  in  the 
figure,  giving  a  reason  in 

fe  ,     '  °         °  Fig.  48. 

each  case. 

2d.  Select  pairs  of  lines  that  you  can  declare  to  be  equal, 
because  of  principles  already  established,  giving  the  prin- 
ciple in  each  case. 

3d.  Select  four  pairs  of  equal  triangles  in  the  figure,  stat- 
ing the  proper  principle  for  each  case. 

4th.  From  the  equal  triangles  show  that  certain  lines  in 
the  figure  must  be  equal  to  each  other  that  were  not  known 
before  to  be  equal.  What  can  you  say  of  the  point  M  ? 
How  do  the  diagonals  divide  each  other  ?  Write  and  re- 
cord a  principle  stating  the  fact  that  you  have  just  learned 
about  the  diagonals  of  a  parallelogram. 

310.  Draw  a  square,  a  rhombus,  and  a  rectangle,  and 
notice  how  their  diagonals  divide  each  other.  Do  the  di- 
agonals ever  cross  at  right  angles  ?  Can  you  account  for 
their  crossing  at  right  angles  by  the  principles  that  enabled 
you  to  draw  a  perpendicular  to  a  line  ? 

311.  Do  the  diagonals  ever  cross  at  a  point  which  is  the 
same  distance  from  all  four  corners  ?  You  have  already 
learned  a  principle  that  should  enable  you  to  answer  this 
question  ;  can  you  recall  it  ? 

312.  What  kinds  of  parallelogram  can  you  inscribe  in  a 


60  ELEMENTARY  AND 

circumference  ?  A  figure  is  inscribed  in  a  circumference 
when  the  circumference  passes  through  all  the  vertices  of 
the  figure. 

312.  On  smooth,  stiff  card-board  draw  two  squares  with 
sides  If  in.  long.  In  the  first  square  draw  one  diagonal.  In 
the  second  square  draw  the  lines  represented  in  the  second 
figure  of  the  plate  on  the  next  page.  Cut  the  squares  into 
seven  pieces,  numbered  as  in  the  diagram. 

Select  all  the  pairs  of  equal  lines  in  the  second  square  ; 
also  give  the  number  of  degrees  in  every  angle  formed. 

313.  With  the  seven  pieces  formed  by  cutting  the  squares 
of  312  a  great  variety  of  interesting  figures  can  be  formed. 
For  instance,  by  placing  them  as  in  the  fourth  figure  of 
the  plate,  an  old-fashioned  chair  like  the  third  figure  of  the 
plate  is  formed. 

314.  Try  to  form  figures  like  those  given  in  the  next 
three  plates.  Each  time  you  must  use  all  seven  pieces.  I 
have  seen  over  three  hundred  figures  formed  with  these 
seven  pieces,  many  of  them  bearing  quaint  resemblances  to 
common  objects.  With  squares  of  wood,  celluloid,  or 
ivory,  divided  in  the  manner  outlined,  you  may  get  much 
amusement  by  attempting  new  forms. 

315.  Arrange  the  seven  pieces  in  such  a  way  as  to  form 
a  rectangle ;  a  rhomboid  ;  a  square.  Notice  the  side  of  the 
square  formed.  Which  of  the  original  lines  forms  its  side  ? 
How  much  larger  is  the  new  square  than  the  old  one  ?  Is  a 
method  of  finding  a  square  twice  as  large  as  a  given  square 
suggested  ? 

316.  If  you  put  pieces  4,  5,  and  7  together  so  that  they 
form  a  square,  how  will  that  square  be  related  to  piece  3  ? 
What  part  of  the  original  square  will  it  be  ? 

317.  Write,  for  review,  short  sketches  of  the  square, 
rhombus,  and  rectangle,  following  as  a  model  the  following 
sketch  of  the  rhomboid  :  The  rhomboid  is  a  quadrilateral 
which  has  two  pairs  of  parallel  sides,  its  adjacent  sides  un- 
equal, and  no  one  of  its  angles  equal  to  90°.  Each  diagonal 
divides  the   rhomboid  into   two  equal  triangles;  the  two 


CONSTRUCTIONAL  GEOMETRY  Gl 


□ 


ELEMENTARY  AND 


y 


LT^ 


^_7 


\ 


CONSTRUCTIONAL  GEOMETRY 


63 


/\ 


N/ 


X 


7 


W 


\ 


7 


Fi<;.  49. 


64  ELEMENTARY  AND 

diagonals  bisect  each  other,  and  divide  the  rhomboid  into 
four  triangles  equal  pair  by  pair.  One  angle  of  a  rhomboid 
determines  all  the  angles,  because  the  opposite  angles  are 
equal,  and  the  adjacent  angles  are  supplementary. 

318.  Draw  a  quadrilateral,  ABCD  (Fig.  49),  by  making 
DCffABand  DC  =  AB.  When  A  D  and  B  C  are 
joined,  will  the  figure  be 
a  parallelogram  ?  Draw 
B  D,  and  see  if  you  can- 
not select  enough  equal 
parts  to  prove  /\ABD 
=  /\  B  D  C.  Arrange 
your  work  neatly,  as  al- 
ready explained  to  you 
in  297.  Draw  as  many 
conclusions  as  you  can  ;  and,  as  a  result  of  your  investiga- 
tion, make  the  following  statement  correct  :  A  quadrilat- 
eral is  or  is  not  necessarily  a  parallelogram,  if  it  has  two  sides 
both  equal  and  parallel. 

319.  Place  two  unequallmes,  M  Pand  N  Q,  with  their  mid- 
dle points  together ;  make  the  lines  cross  at  an  oblique  angle. 
Can  you  show  what  kind  of  quadrilateral  MNPQ  must 
be  ?     Record  the  principle,  when  you  have  discovered  it. 

320.  Could  you  so  place  the  lines  M  P  and  N  Q  of  319 
that  the  figure  MNPQ  would  be  a  rhombus  ? 

321.  If  you  wish  to  place  two  sticks  so  as  to  make  a 
rectangular  kite,  what  conditions  must  the  sticks  fulfil  ? 
Would  the  same  conditions  enable  you  to  place  the  sticks 
so  as  to  make  a  square  kite  ? 

322.  Make  a  quadrilateral,  A  B  C  D,  by  making  A  B  = 
D  C  and  A  D  =  B  C.  Will  the  quadrilateral  have  any  pecul- 
iar form  ?  Can  you  prove  it  by  drawing,  as  a  help  line,  the 
diagonal  B  D  ?    As  a  result   of  your  investigation  complete 

the  following  sentence  :  A  quadrilateral be  a if 

it  has  its  opposite  sides  equal  pair  by  pair 

323.  Can  you  draw  a  quadrilateral  with  two  pairs  of 
equal  sides  that  does  not  come  under  the  principle  of  322  ? 


CONSTRUCTIONAL   GEOMETRY  65 


SECTION  XIII.     DIVISION  OF  LINES. 

324.  Draw  a  scalene  triangle,  ABC 
scalene  triangle  is  one  that  has 
its  sides  unequal.  A  scalene 
triangle  bears  the  same  relation 
to  triangles  that  a  rhomboid  does 
to  parallelograms.  It  is  the 
kind  of  triangle  that  should  be 
drawn  for  illustration  when  no 
particular  kind  is  specified. 

Find  M,  the  middle  of  A  C, 
and  draw  M  N  symparallel  with 
A  B.  (Let  your  construction 
lines  be  light  but  visible.) 

Draw  a  help  line  N  K  f  f  MA.  (Note  that  help  lines 
are  always  dotted  lines.) 

Study  the  figure  thus  formed,  and  write  answers  to  the 
following  questions,  giving  reasons  in  each  case  : 

How  many  and  what  angles  are  equal  to  /  A  ? 

What  angle  is  equal  to  /  C  ?  to  /  B  ? 

What  kind  of  quadrilateral  is  A  K  N  M  ? 

What  lines  can  be  inferred  to  be  equal  from  the  nature 
of  the  quadrilateral  A  K  N  M  ? 

How  many  angles  of  A  M  N  C  can  be  found  mated  in 
AKBN? 

Is  any  side  of  A  M  N  C  known  to  be  equal  to  a  side  of 
A K  B  N  ? 

Is  A  M  N  C  equal  to  A  K  B  N  ? 

If  you  have  been  able  to  answer  these  questions,  and  to 
support  your  answers  with  good  reasons,  you  have  proved 
that  C  N  =  N  B  and  M  N  =  K  B  ;  so  that  N  is  the  middle 
point  of  B  C,  and  M  N  is  one-half  of  the  base  line  A  B. 

325.  The  principle  involved  in  324  is  one  of  the  most 
important  in  Geometry.  Stated  in  general  language  it  is 
as  follows  : 

5 


B 


66  ELEMENTARY  AND 

A  line  drawn  from  the  middle  point  of  one  side  of  a  triangle 
parallel  to  the  base  bisects  the  other  side,  and  is  itself  one- 
half  of  the  base. 

326.  This  principle  suggests  a  new  and  simple  method  of 
bisecting  lines  by  using  common  ruled  paper,  where  the  lines 
are  drawn  equal  distances  apart.  A  sheet  of  theme-paper 
is  excellent  for  this  purpose,  because  there  is  a  red  line  at 
right  angles  to  the  blue  ruled  lines,  and  because  the  paper 
is  broad.  Along  the  red  line  number  successive  lines,  0,  1, 
2,  3,  etc.  (Fig.  51).  Place  the  line  to  be  bisected  on  line 
2  with  one  end  at  point  2 ;  con- 
nect the  other  end,  A,  with  point 
0  by  your  ruler's  edge ;  1  B  is 
clearly  £  of  2  A.  The  line  0  A 
need  not  be  drawn ;  the  same 
piece  of  paper,  therefore,  may  be 
used  for  bisecting  an  indefinite 
number  of  lines. 

327.  On  what  other  lines  be- 
tween 0  and  12  could  you  place 
the  line  to  be  bisected,  and  on 

what  lines  respectively  would  the  half-line  appear  ?  Bisect 
a  line  by  placing  it  on  three  different  lines,  and  compare 
your  results.  Can  you  see  any  principle  to  guide  your 
choice  of  a  line  on  which  to  place  the  line  to  be  bisects  I  'i 

328.  Make  a  triangle,  and  then  make  another  with  its 
sides  one-half  as  long,  using  the  new  method  for  getting 
the  half-lines. 

329.  Make   a   quadrilateral — not   a  parallelogram — and 
find  the  middle  points  of  its  sides  by  the  method  of 
Join  the  middle  points  of  the  sides  in  order  by  straight  lines. 
What  kind  of  quadrilateral  will  be  formed  ?     Can  you  ac- 
count for  its  shape  ? 

Suggestion  :  Draw  one  diagonal  of  the  original  quadri- 
lateral, and  apply  what  you  learned  in  325  and  318.  The 
principle  of  325  does  not  apply  immediately,  but  it  leads 
very  easily  to  the  right  principle. 


CONSTRUCTIONAL   GEOMETRY  67 

330.  Make  a  rhomboid,  and  repeat  the  experiment  of 
329.  Can  you  account  for  the  shape  of  the  new  parallelo- 
gram formed  ?     Can  it  be  anything  but  a  rhomboid  ? 

331.  Repeat  the  experiment  of  329,  using  a  rectangle. 
What  kind  of  parallelogram  will  be  formed  by  joining  the 
middle  points  of  the  sides  ? 

332.  Try  the  same  experiment,  using  a  rhombus  at  first. 
Account  for  the  shape  of  the  second  parallelogram  formed. 

333.  Draw    a  trapezoid,  A  B  C  D  (Fig.  52).     Find  M, 
the  middle  point  of  AD, 
draw  M  N  f  f  A  B,  and  study 
the  figure  formed,  using  as  a 


help  line  the  diagonal  D  B.        m/- -^ 

What  reason  have  you  for 
thinking  K  the  middle  of 
DB?  A  ^ 

Compare  M  K  with  A  B.  FlG- 52- 

Have  you  any  reason  for  thinking  N"  to  be  the  middle  point 
of  B  C  ?  Suggestion  :  Study  the  ^  D  B  C.  Compare  K  N 
with  D  C.  Compare  M  N  with  A  B  and  D  C  combined.  The 
line  MNisa  famous  line  in  the  trapezoid.  It  is  called  the 
median  of  a  trapezoid  because  it  passes  though  the  middle 
points  of  both  legs.  Notice  carefully  how  the  median  was 
drawn.  It  was  not  drawn  from  middle  point  to  middle  point, 
but  from  one  middle  point  parallel  to  the  two  bases,  and  was 
proved  to  pass  through  the  middle  point  of  the  second  leg,  a 
characteristic  from  which  it  gets  its  name.  Perhaps  a  more 
important  characteristic  is  that  it  is  half  of  the  sum  of  the 
bases,  or  the  average  of  the  two  bases.  It  is  very  important 
that  you  should  become  familiar  with  both  characteristics 
and  with  the  following  principle  :  The  line  drawn  from  the 
middle  point  of  one  leg  of  a  trapezoid  parallel  to  the  bases  bi- 
sects the  other  leg,  and  is  itself  one-half  the  sum  of  the  bases. 

334.  Make  two  lines,  a  and  b,  and  find  their  average  by 
the  principle  of  333,  and  with  the  aid  of  the  ruled  sheet  of 
paper  used  in  326  ;  also  add  the  two  lines  a  and  b,  and  find 
one-half  of  the  resulting  line  by  326.  Compare  the  results. 


OF  THE 

(    UNIVERSITY 


68  ELEMENTARY  AND 

335.  What  is  the  average  width  of  a  board  which  is  12 
in.  at  one  end  and  9  in.  at  the  other  ?  Do  this  both  by 
Geometry  and  by  Arithmetic,  and  compare  the  results. 

336.  Find  the  average  diameter  of  an  ordinary  wooden 
pail.  Can  you  find  the  average  diameter  of  a  barrel  ?  of 
a  base-ball  bat  ? 

337.  In  a  trapezoid,  ABCD,  find  the  median,  if  A  B  = 
11|  in.  and  C  D  =  5 J  in. 

338.  Find  the  base  A  B  of  a  trapezoid,  A  B  C  D,  if  C  D 
=  8  in.,  and  the  median,  M  N,  =12  in.;  if  C  D  =  11  in. 
and  the  median,  MN,  =  13 \  in.;  if  C  D  =  7£  cm.,  and 
M  N  =  4|  cm. 

339.  On  a  line,  A  D  (Fig.  53),  mark  off  five  equal 
lengths  with  your  compasses.  Draw  A  B  making  an 
oblique  angle  with  A  D,  and  through  the  points  of  divis- 
ion on  A  D  draw  parallels  to  A  B  terminating  in  a  line 
B  C.  Measure  the  upper  two  parallels  and  calculate  the 
rest. 

340.  Could  you  calculate  all  the  lines  from  measurements 
of  the  first  and  third  lines, 
D  C  and  E  L?  of  the  first  and 
fourth  lines,  D  C  and  G  M  ? 

341.  On  a  sheet  of  ruled 
paper  draw  two  non-parallel 
lines  in  such  a  way  that  they 
shall  intercept  \  in.  on  the 
top  line  and  1  in.  on  the  line 
below  the  top.  Calculate  the 
lengths  intercepted  on  all  the 
lines.  "  B 

342.  On  a  piece  of  theme- 
paper  mark  the  points  at  which  the  red  line  crosses  the  blue 
lines,  0,  1,  2,  3,  4,  5,  etc.,  in  succession  ;  draw  from  the 
point  0  a  line,  0  Y,  crossing  obliquely  the  blue  lines  at 
points  A,  B,  C,  D,  E,  etc.,  respectively.  If  you  rule  the 
paper  yourselves,  take  pains  to  have  the  lines  drawn  at 
equal  distances  and  strictly  parallel.      Study  the  triangle 


CONSTRUCTIONAL  GEOMETRY  69 

0  2  B.  What  can  you  say  about  the  line  1  A  ?  the  line  0  A  ? 
If  you  give  the  name  x  to  the  line  1  A,  what  name  will  you 
give  to  the  line  2  B  ? 

In  the  next  place  study  the  trapezoid  1 A  C  3.     The  line 

2  B  is  manifestly  the  median  of  the  trapezoid,  and,  there- 
fore, the  average  of  the  two  bases.     It  follows  that  the  line 

3  C  must  be  equal  to  3x,  for  2x  is  the  average  between  x 
and  3x.  By  studying  the  different  trapezoids  in  order, 
you  will  see  that  4  D  is  four  times  as  long  as  1  A,  or  \x  ; 
that  5  E  is  five  times  as  long  as  1  A,  or  5x,  etc.  What 
would  be  the  distance  cut  off  on  line  number  10  by  0  Y  ? 
on  line  number  16  ?  Wherever  you  can,  confirm  your  re- 
sults by  the  middle-point  principle  of  326  and  327.  Thus 
by  studying  /\04  D  you  see  that  4  D  ought  to  be  twice 
as  long  as  2  B,  or  4x,  the  result  already  obtained. 

343.  Does  it  make  any  difference  in  the  relative  value  of 
the  lines  cut  off  whether  0  Y  diverges  more  or  less  from 
the  red  vertical  line  ? 

344.  If  you  should  place  a  line  one  inch  long  on  line 
number  10 — placing  one  end  of  the  inch  line  on  the  red 
line  and  joining  the  other  end  with  0 — what  part  of  an 
inch  would  be  found  on  line  No.  1  ?  on  line  No.  3  ?  Have 
you  ever  seen  a  ruler  for  getting  tenths  of  an  inch  arranged 
on  this  principle  ?  Cannot  some  one  of  the  class  obtain 
such  a  ruler  for  illustration  in  the  class-room  ? 

345.  How  would  you  use  these  same  ten  lines  to  obtain 
jfo  of  an  inch  ?    Tf 7  of  an  inch  ? 

346.  Compare  the  line  3  C,  on  line  No.  3,  with  the  line  5  E 
on  line  No.  5  ;  also  compare  line  4  D  with  line  7  G.  Then 
explain  how  you  would  get  §  of  a  line,  $  of  a  line,  f  of  a 
line,  TV  of  a  line  by  means  of  a  sheet  of  ruled  theme-paper. 

347.  Place  any  line,  P  Q,  on  the  tenth  line, with  P  at  point 
10.  What  part  of  P  Q  will  be  found  between  your  ruler 
and  point  7,  if  you  place  your  ruler's  edge  from  0  to  Q  ? 

348.  Can  you  get  J  of  a  line  by  using  the  same  sheet  of 
paper  ?  Explain  your  method.  Draw  a  line,  P  Q,  and  get 
|  of  it ;  %  of  it ;  2 J  of  it. 


70 


ELEMENTARY  AND 


349.  Make  a  trefoil  (Fig.  11),  with  the  centres  about  one 
inch  apart.  Then  make  a  figure  just  like  it  in  shape 
—or  similar  to  it— with  all  the  lines  $  as  long.  Then 
enlarge  the  original  figure  by  making  the  lines  }  as  long. 

350.  Make  a  rose  window  (Fig.  54)  by  putting  the  cen- 
tres of  the  arches  at  points  on 
a  £  in.  circumference  which  are 
f  of  an  inch  apart.  Reduce  the 
figure  by  making  the  lines  ^  as 
long ;  also  enlarge  the  figure  by 
making  the  lines  £  as  long. 

351.  With  only  ten  ruled  lines 
can  you  find  -fc  of  a  line?  &  of 
aline  ?  TJT  of  aline  ? 

352.  Can  you  find  $  of  a  line 
too  long  to  put  on  your  paper  ?  r,°  M- 

353.  Find  a  line  -^  as  long  as  your  desk. 

354.  Can  you  find  fT  of  a  line  with  a  sheet  of  paper  con- 
taining only  ten  parallel  lines  ? 

355.  It  will  be  convenient  for  the  future  to  refer  to  your 
ruled  sheet  of  theme-paper  as  your  dividing  tool;  for  with 
it  you  can  get  almost  any  fraction  of  a  line  that  you  wish. 

356.  If,  instead  of  putting  point  0  on  the  top  line,  we  had 
put  it  on  line  5,  say,  and  about  half-way  between  the  ed| 
and  if  we  had  numbered  the  lines  above  this  zero  line, 
1,  2,  3,  4,  5,  and  the  numbers  below  the  zero  line,  1,  2, 
3,  4  5,  6,  7,  etc.;  and  if  we  had  drawn  two  oblique  lines 
meeting  at  0  across  all  the  lines,  we  should  have  had  a 
picture  of  an  instrument  called  proportional  dividers,  a 
very  useful  instrument  in  reducing  or  enlarging  a  figure. 
The  principle  is  the  same  as  that  of  your  dividing  tool,  but 
the  tool  is  more  convenient  than  yours  in  actual  practice. 
Any  line  between  the  oblique  lines  on  line  5  above  the  zero 
line  is  the  same  in  length  as  the  corresponding  line  on  line 
5  below  the  zero  line,  and  this  latter  line  we  know  to  be  f 
as  long  as  the  length  intercepted  on  line  7  by  the  oblique 
lines.     Make  a  diagram  to  illustrate  what  has  been  said, 


CONSTRUCTIONAL   GEOMETRY  71 

and  see  if  you  can  follow  it.  Proportional  dividers  are  di- 
viders pivoted,  not  at  one  end  as  your  dividers  are,  but  some- 
where between  the  ends  at  a  point,  corresponding  to  0,  that 
can  be  changed  at  pleasure  by  a  sliding  screw.  If  it  is 
wished  to  reduce  the  lines  of  a  figure  to  $  of  their  actual 
size,  the  dividers  are  pivoted  so  that  the  legs  on  one  side  of 
the  pivot  are  f  as  long  as  those  on  the  other  side.  Now, 
when  the  longer  legs  are  opened  any  distance,  the  shorter 
legs  are  opened  only  f  as  far  ;  so  that  all  the  reduced  lines 
can  be  found  with  great  rapidity  without  changing  the  pivot 
screw.  The  figures,  pp.  61-63,  were  reduced  from  actual 
blocks  by  proportional  dividers.  If  possible,  secure  a  pair 
of  proportional  dividers  to  illustrate  the  important  principle 
involved. 


REVIEW. 

357.  Review  from  281,  and  come  into  the  class  prepared 
to  write  a  short  sketch  of  any  of  the  quadrilaterals  thus 
far  studied. 

358.  Do  the  diagonals  of  a  trapezoid  or  a  trapezium  ever 
bisect  each  other  ? 

359.  What  can  be  said  of  the  diagonals  of  a  rhombus  that 
cannot  be  said  of  the  diagonals  of  a  rectangle,  and  vice  versa  f 

360.  A  quadrilateral  when  cut  along  one  diagonal,  was 
found  to  be  divided  into  two  equal  pieces ;  can  you  say 
what  the  shape  of  the  quadrilateral  was  ?     Why  ? 

361.  Two  sides  of  a  quadrilateral  were  found  to  be  equal 
and  to  be  symparallel ;  what  possible  shapes  could  the  fig- 
ure have  had  ? 

362.  In  studying  a  certain  quadrilateral  a  pupil  noticed 
that  the  point  where  the  diagonals  crossed  was  the  same  dis- 
tance from  all  the  corners  ;  do  you  know  what  quadrilateral 
he  was  studying  ? 

363.  If  you  have  two  lines  of  unequal  length,  how  will 
your  dividing  tool  enable  you  to  tell  whether  the  shorter  is 
more  or  less  than  T\  of  the  longer  ? 


72  ELEMENTARY  AND 

364.  What  principle  of  the  trapezoid  leads  to  the  con- 
struction of  your  dividing  tool  ? 

365.  With  a  pair  of  proportional  dividers  how  would  you 
reduce  the  figures  on  the  plates  following  314  so  that  the 
lines  would  be  §  as  long  ? 

366.  If  on  side  A  B  of  /\  A  B  C  a  point,  P,  is  chosen 
I  of  the  distance  from  C  to  A,  and  if  from  Pa  line,  P  Q,  is 
drawn  symparallel  to  the  base  A  B,  what  part  of  A  B  will 
P  Q  be  ?    What  part  of  C  B  will  C  Q  be  ? 


SECTION  XIV.   MULTIPLICATION  OF  LINES. 
AREAS. 

367.  We  have  already  learned  how  to  add  lines  and  to 
subtract  lines  ;  in  certain  cases,  with  the  aid  of  the  dividing 
tool,  we  can  divide  one  line  by  another  ;  for  the  line  inter- 
cepted on  the  ninth  parallel  by  0  Y  of  the  dividing  tool  will 
contain  the  line  intercepted  on  the  fourth  parallel  by  0  Y  2^ 
times.  The  question  naturally  arises  whether  or  not  two 
lines  can  be  multiplied  together. 

Draw  a  line,  M  N  (Fig.  55),  and  with  your  dividing  tool 
find  M  Q  =  {  M  N.  Place  the  line  M  Q  at  right  angles  to 
M  N,  and  complete  the  rectangle  M  N  S  Q.  Now  divide 
M  N  into  nine  equal  parts  with  your  dividing  tool.  How 
many  of  these  parts  will  M  Q  contain  ?  Why  ?  Through 
the  points  of  division  in  M  N  draw  parallels  to  M  Q,  and 
through  the  points  of  division  in  M  Q  draw  parallels  to 
M  N.  The  small  figures  formed  will  be  squares.  Why  ? 
How  many  of  them  ? 

How  many  squares  would  there  have  been,  had  you  made 
M  Q  =  £  of  MN?  How  many  squares  in  the  figure 
MTRQ?  If  each  one  of  the  divisions  of  M  N  is  called  "  b," 
each  one  of  the  little  squares  is  called  a  "  square  b  "  or  "  b 
square "  (written  b2).    It  is  easy  to  see  from  experiments 


CONSTRUCTIONAL  GEOMETRY 


73 


with  different  rectangles  in  Fig.  55  that  any  rectangle  con- 
tains as  many  "  square  b's  "  as  there  are  units  in  the  product 
obtained  by  multiplying  the  number  of  b's  in  the  base  by  the 
number  of  b's  in  the  side.  Thus  the  rectangle  TNSR, 
with  two  b's  in  the  base  and  five  b's  in  the  side,  contains 
two  lines  five  or  ten  "  square  b's,"  written  1062. 

In  describing  a  floor  we  say  that  it  contains  300  square 
feet — meaning  by  that  that  it  covers  as  much  surface  as 


h 


Mhr 


N 


Fig.  55. 


300  squares  each  1  ft.  long  and  1  ft.  wide  would  cover ;  and 
we  find  out  how  many  square  feet  it  covers  by  measuring 
the  length  and  width  in  feet,  and  by  multiplying  the  two 
results  together.  In  the  same  way  we  measure  M  N  and 
M  Q  in  b's,  and  multiply  the  number  of  b's  in  M  N  by  the 
number  of  b's  in  M  Q,  thus  obtaining  the  number  of 
u  square  b's  "  covered  by  the  rectangle  MNSQ. 

368.  It  is,  therefore,  a  simple  step  to  say  that  the  product 
of  two  lines  represents  a  surface,  bounded  by  a  rectangle  of 
which  the  two  lines  are  adjacent  sides.  To  multiply  two 
lines,  then,  according  to  this  understanding  of  the  terms, 
we  simply  complete  a  rectangle  with  the  two  lines  as  ad- 
jacent sides.     It  is  not  necessary  to  form  each  time  a  pict- 


74  ELEMENTARY  AND 

ure  of  the  little  squares  that  correspond  to  b2  in  Fig.  55. 
On  the  contrary,  it  is  better  to  think  of  the  rectangular  sur- 
face as  a  whole  in  thinking  what  the  product  of  two  lines 
means. 

369.  Draw  three  pairs  of  lines,  and  multiply  the  lines  of 
each  pair  together.  It  will  be  well  to  review  the  method  of 
drawing  a  perpendicular  at  the  end  of  a  line  given  in  8 ■.'.".. 

370.  Draw  three  rectangles,  and  tell  what  lines  must  be 
multiplied  together  to  produce  them. 

371.  In  problems  relating  to  the  surface,  or  area,  of  rec- 
tangles, the  upright  side  is  called  the  altitude  of  the  rectan- 
gle, and  the  line  on  which  the  rectangle  is  supposed  to 
stand  is  called  the  base.  The  principle  stated  in  the  pic- 
ceding  articles  of  this  section  may  be  expressed  as  follows  : 
The  product  of  the  base  and  the  altitude  of  a  rectangle  is  the 
surface  of  the  rectangle. 

372.  Although  the  simplest  product  of  two  lines  is  a  rec- 
tangular surface,  fortunately  we  are  not  limited  to  this  sur- 
face. 

Draw  a  rhomboid  A  B  C  D  (Fig.  56  (a)),  and  draw 
B  H  1  A  B.  Cut  the  /\  B  II C  from  the  right  of  Fig.  56  (a) 


(a.)  FlQ   66.  (6.) 

and  place  it  to  the  left  of  A  D,  as  in  Fig.  56  (b).  How 
do  you  know  that  B  C  will  fit  on  A  D,  and  that  D  K  will 
be  in  a  straight  line  with  II  D  ?  How  does  the  surface  of 
the  rectangle  ABHK  compare  with  the  surface  of  the 
rhomboid  A  B  C  D  ?  What  product  represents  the  rectan- 
gle ?  What  product  represents  the  rhomboid  therefore  ? 
Illustrate  this  article  by  cutting  a  rhomboid  out  of  paper. 


CONSTRUCTIONAL  GEOMETRY 


75 


373.  The  surface,  or  area,  of  the  rhomboid  is  the  same 
as  the  surface  of  the  rectangle ;  but  the  rhomboid  is  not 
equal  to  the  rectangle  because  one  figure  would  not  fit  the 
other  exactly.  When  two  areas  are  the  same  the  figures 
are  said  to  be  equivalent ;  a  new  sign,  =o,  is  very  com- 
monly used  to  express  this  condition,  although  it  is  not  a 
sign  universally  agreed  upon.  The  sentence  /  J  ABC 
D  O  1  |  A  B  H  K  is  translated  thus  :  The  rhomboid 
A  B  0  D  is  equivalent  to  the  rectangle  ABHK;  and 
the  meaning  is  that  the  figures  have  the  same  area. 

374.  What  is  the  altitude  of  the  rhomboid  A  B  C  D, 
Fig.  56  (a)  ?  Translate  this  sentence  :  /  7  A  B  C  D  =o= 
A  B  x  B  H. 

375.  Make  four  rhomboids  of  different  shapes,  and  draw 
the  lines  whose  product  gives  the  area  in  each  case,  record- 
ing the  result  according  to  the  geometrical  sentence  in  374. 
Is  there  more  than  one  answer  in  each  case  ? 

376.  Multiply  two  lines  so  that  the  product  shall  be  a 
rhomboid.  Can  you  get  more  than  one  rhomboid  for  an 
answer  ?     How  many  ? 

377.  Can  you  multiply  two  lines  and  obtain  the  result- 
ing rhomboid  at  once,  without  first  drawing  the  rectangle 
of  the  two  lines  ? 


378.  Make  two  parallel  lines,  A  F  and  K  P  (Fig.  57). 
Make  AB  =  OD  =  EF,  and  complete  the  rectangle 
A  B  H  K  and  the  two  rhomboids  0  D  M  N  and  E  F  P  0. 
Which  of  the  three  parallelograms  has  the  largest  surface  ? 
Why  ?     Make  clear  what  product  represents  each  surface. 


76 


ELEMENTARY  AND 


379.  The  sum  of  the  bounding  lines  of  a  figure  is  OoDed 
its  perimeter.  Do  the  three  parallelograms  of  Fig.  57 
have  equal  perimeters  ? 

380.  If  a  man  has  an  acre  of  ground,  will  it  make  any 
difference  to  him,  in  the  cost  of  fencing  it,  whether  the 
field  is  in  the  shape  of  a  rectangle  or  of  a  rhomboid  ? 
Which  is  the  more  economical  shape  ? 

381.  Can  you  arrange  a  square  foot  of  paper  in  the  form 
of  a  rhomboid  with  a  perimeter  of  fifty  feet  ?  Can  you 
make  the  perimeter  as  long  as  you  please  without  changing 
the  surface  of  the  paper  ?  as  short  as  you  please  ? 

382.  In  building  a  cistern  of  a  fixed  height  to  hold  a  cer- 
tain amount  of  water,  does  it  make  any  difference  in  the 
number  of  bricks  used  whether  you  make  the  bottom  of 
the  cistern  in  the  shape  of  a  rhomboid  or  of  a  rectangle  ? 

383.  Which  lets  in  the  more  light,  a  diamond-shaped 
window  pane  or  a  rectangular  window  pane,  if  the  perim- 
eters of  the  two  panes  are  the  same  ? 

Make  two  lines,  a  and  b.  Make  a  rhomboid  equivalent 
to  a  x  b  which  shall  have  a  perimeter  three  times  the  sum 
of  a  and  b. 

384.  If  a  farmer  wants  to  have  as  little  fencing  as  possi- 
ble to  build,  and  if  he  is  obliged  to  have  his  field  in  the 
shape  of  some  parallelogram, 
what  parallelogram  ought  he 
to  choose  ? 

385.  Under  what  conditions 
are  two  parallelograms  with 
the  same  base  equivalent  ?  Il- 
lustrate your  answer,  and  re- 
cord the  principle  involved. 

386.  Can  the  product  of  two 
lines  ever  be  a  triangular  sur- 
face ?  This  is  merely  another 
way  of  asking  whether  a  tri- 
angular surface  can  be  changed  into  an  equivalent  rectan- 
gular surface.     Draw  a  scalene  triangle,  ABC  (Fig.  58). 


CONSTRUCTIONAL   GEOMETRY  77 

Find  the  middle  point,  M,  of  A  C,  and  draw  M  N  parallel 
to  A  B.  Draw  from  C  a  line,  C  H  1  A  B.  (Show  your 
construction  lines. )  What  is  C  H  called  ?  What  part  of 
C  His  CD?    Why? 

Erect  perpendiculars  to  A  B  at  A  and  B.  Extend  D  M 
and  D  N  until  they  meet  these  lines  at  K  and  S  respec- 
tively. Show  that  A  0  D  N  =  A  S  K  B  and  that  AC  M  D  = 
AMKA. 

IsAABC=o=[^]ABSK?    Why? 

By  what  line  must  you  multiply  A  B  to  produce 
A  ABC? 

Follow  the  steps  of  this  article  with  a  paper  triangle 
which  you  cut  along  the  lines  indicated. 

387.  Draw  five  different  triangles,  and  draw  side  by  side 
the  lines  whose  product  in  each  case  represents  the  area  of 
the  triangle.     In  two  of  the  cases  make  a  base  angle  obtuse. 

388.  Make  three  pairs  of  unequal  lines.  Form  three  tri- 
angles whose  surfaces  will  be  represented  by  the  products  of 
the  lines  respectively  taken  pair  by  pair. 

389.  Is  there  only  one  triangle  to  be  found  equivalent 
to  the  product  of  two  lines,  a  and  b  ?  If  more  than  one, 
how  many  can  be  found  ?      Do  their  perimeters  differ  ? 

390.  Make  two  parallel  lines ;  on  the  lower  line,  A  D, 


Fig.  59. 


take  three  equal  lengths,  A  B,  E  F,  and  C  D  ;  on  the  up- 
per line  take  three  points,  P,  R,  and  Q,  at  random  ;  com- 
plete the  triangles   A  B  P,  E  F  R,  and  C  D  Q.     Which 
has  the  largest  area  ?     Which  has  the  longest  perimeter  ? 
If  you  wish  to  find  the  triangle  which  has  the  shortest 


78  ELEMENTARY  AND 

perimeter  possible,  provided  that  the  base  equals  A  B  and 
that  the  vertex  is  on  P  Q,  what  shape  will  you  give  the  tri- 
angle ? 

391.  Pin  the  ends  of  a  short  piece  of  elastic  to  a  line  A  B 
at  the  points  A  and  B  ;  then  form  a  triangle  A  B  P  with 
the  elastic  without  stretching  the  elastic ;  draw  through 
P  a  line,  P  Q,  parallel  to  A  B,  and  pull  the  vertex  P  of  the 
elastic  triangle  along  P  Q.  In  this  way  you  can  mould  the 
triangle  A  B  P  into  a  triangle  having  almost  any  desired 
shape.  Does  the  area  change  in  the  process  ?  When  is 
the  perimeter  shortest  ?  If  the  elastic  could  be  stretched 
indefinitely,  would  there  be  any  limit  to  the  perimeter  ? 

392.  Can  you  mould  the  elastic  triangle  of  391  into  a 
right  triangle  ?  an  isosceles  triangle  ?  an  equilateral  tri- 
angle ? 

393.  Learn  by  heart  the  following  principle  : 

If  the  vertex  of  a  triangle  is  moved  parallel  to  the  base,  the 
area  of  the  triangle  remains  unchanged  as  long  as  the  base 
remains  unchanged. 

394.  Make  a  triangle,  ABC,  and  mould  it  into  an  equiv- 
alent triangle  with  one  side  twice  as  long  as  A  C  without 
changing  the  base  A  B. 

395.  Make  a  triangle,  ABC,  and  mould  it  into  an  equiv- 
alent isosceles  triangle ;  mould  it  also  into  a  triangle  with 
base  angle  70°. 

396.  Make  two  lines,  a  and  b,  and  find  the  triangle 
represented  by  their  product,  without  constructing  an  aux- 
iliary rectangle.  Repeat  the  problem  with  three  different 
pairs  of  lines. 

Note. — The  problem  should  be  repeated  until  it  is  as  easy 
to  form  the  "  triangle  of  two  lines  "  as  it  is  to  form  the 
"rectangle  of  two  lines." 

397.  Make  a  pyramid  with  a  square  base.  Draw  the 
lines,  five  pairs,  the  sum  of  whose  products,  pair  by  pair, 
represents  the  total  surface  of  the  pyramid. 

398.  If  three  triangular  tin  boxes  of  equal  height,  with 
bottoms  and  tops  shaped  like  the  three  triangles  of  Fig.  59, 


CONSTRUCTIONAL  GEOMETRY 


79 


were  to  be  made,  which  box  would  hold  the  most  sand  ? 
Would  one  box  require  more  tin  than  another  ?  Why  ? 
Can  you  draw  pictures  of  the  boxes  ?  What  is  the  geomet- 
rical name  for  the  sides  of  the  boxes  ?  When  boxes  are  made 
with  triangular  bottoms,  what  shape  is  usually  chosen  ? 
Do  you  understand  why  ? 


SECTION  XV.     EQUIVALENT   FIGURES. 
MOULDING   OF  AREAS. 

399.  The  principle  of  393  enables  you  to  mould  triangles 
and  all  plane  figures  of  geometry  almost  as  easily  as  if  the 
figures  were  made  of  elastic  or  of  clay.  For  instance,  sup- 
pose that  two  triangles,  A  A  B  C  and  A  B  D  E  of  Fig.  60, 
are  of  the  same  height,  and  are  standing,  side  by  side,  on 
the  same  base  line  as  in  Fig.  60.     They  can  be  moulded 


Fig.  60. 

into  one  triangle,  A  C  D,  with  perfect  ease.     Do  you  see 
why  A  B  C  D  is  equivalent  to  A  B  E  D  ? 

Note. — In  this  figure  and  in  future  figures,  when  figures 
are  moulded,  the  lines  along  which  vertices  are  moved 
will  be  indicated  by  light  dotted  lines,  as  all  help  lines 
should  be  indicated  ;  the  resulting  lines  of  the  remodelled 
figure  will  be  indicated  by  double  dotted  lines.     In  your 


80  ELEMENTARY  AND 

drawings  follow  the  same  practice,  unless  you  are  able  to 
use  different  colors  for  the  different  kinds  of  lines. 

400.  Add  together  two  triangles  which  have  the  same 
altitude,  but  different  bases.  Repeat  with  three  pairs  of 
triangles. 

401.  Can  you  add  three  triangles  that  have  the  same 
altitude  ?  Can  you  do  this  problem  with  only  one  double 
dotted  line  ? 

402.  Find  an  isosceles  triangle  equivalent  to  three  tri- 
angles that  have  the  same  altitude. 

403.  Mould  three  triangles  having  the  same  altitude  into 
a  right  triangle. 

404.  Make  six  equilateral  triangles  from  one  point,  O, 
forming  the  six-sided  figure,  or  hexagon,  represented  in 
Fig.     61.     With    a    sharp 

knife  cut  the  figure  along 
the  lines  O  A,  O  B,  etc., 
beginning  at  O  and  not  cut- 
ting quite  up  to  points  A, 
B,  C,  etc.,  except  at  point 
F.  Place  the  figure  thus 
formed  so  that  A,  B,  C,  D, 
E,  F  will  form  a  straight 
line.  Can  you  form  one 
triangle  as  large  as  the 
hexagon  ?  IIow  long  will 
its  base  be  ?    How  tall  will  *io.  « 

it  be? 

405.  Could  you  mould  a  square  into  a  triangle  by  a  pro- 
cess similar  to  that  of  404  ?  a  parallelogram  ? 

406.  Make  a  quadrilateral,  A  B  C  D,  Fig.  62.  The 
problem  is  to  mould  this  quadrilateral  into  an  equivalent 
triangle.  The  principle  of  393  suggests  a  method.  Di\  Mta 
A  B  C  D  into  two  triangles  by  the  diagonal  B  D.  C  D  B 
can  now  be  looked  at  as  a  triangle  standing  on  the  base  D  B 
with  vertex  at  C.  C  can  be  moved  parallel  to  D  B  as  far 
as  you  wish  to  move  it,  and  D  C  and  B  C  will  follow  the 


CONSTRUCTIONAL  GEOMETRY  81 

moving  C  just  as  the  legs  of  the  elastic  triangle  in  391 
followed  the  moving  vertex.  When  C  reaches  R,  in  the 
line  of  A  B,  /\  C  D  B  has  been  moulded  into  /\RDB,  and 
the  quadrilateral  A  B  C  D  has  been  moulded  into  /^ARD. 


Fio.  62. 

This  process  can  be  illustrated  very  prettily  on  a  moulding- 
tray  with  clay  or  sand  in  the  shape  of  a  quadrilateral,  and 
should  be  so  illustrated  if  possible. 

407.  If  you  had  moved  C  in  a  line  antiparallel  to  D  B, 
where  would  you  have  stopped  the  movement  ?  Illustrate 
with  a  figure,  using  the  same  quadrilateral  as  in  406. 

408.  Draw  the  diagonal  C  A,  Fig.  62,  and  mould  the 
quadrilateral  into  a  triangle  in  two  ways,  first  moving  D 
symparallel  with  C  A,  and  in  a  second  figure  moving  D 
antiparallel  to  C  A. 

409.  Make  a  quadrilateral,  and  mould  it  into  an  isosceles 
triangle. 

410.  Make  a  quadrilateral,  and  mould  it  into  a  right  tri- 
angle. 

411.  Can  you  reverse  the  process  of  406,  and  mould  a  tri- 
angle ADR  (Fig.  62)  into  a  quadrilateral  A  B  C  D  ?  Sug- 
gestion :  Choose  any  point  between  A  and  R  for  B,  and  draw 
D  B  as  a  guide  line,  to  which  you  must  make  R  C  parallel. 
If  all  the  lines  and  points  of  Fig.  62  were  blotted  out  ex- 
cept /\ARD,  what  would  be  your  chances  of  reproducing 
quadrilateral  A  B  C  D  ? 

6 


82  ELEMENTARY  AND 

412.  Make  a^ABC  and  mould  it  into  a  trapezoid. 
What  difficulty  can  you  see  in  moulding  a  triangle  into  a 
parallelogram  by  this  process  ? 

413.  In  moulding  a  triangle,  ABC,  into  a  quadrilateral 
(Fig.  63),  if,  after  choosing  the  point  P  between  A  and  B  for 
one  vertex  of  your  quadrilateral,  you  should  move  B  parallel 
to  P  C,  not  stopping  until  you 
should  reach  A  C  extended, 
the  figure  formed  would  be  a 
new  triangle,  A  P  Q.  The 
base  of  ^  A  B  C  has  been 
shortened  and  the  altitude 
lengthened.  Hence  it  is  pos- 
sible to  shorten  the  base  of  a 
triangle  to  any  desired  length. 
Illustrate  this  article  with  a 
clay  or  sand  triangle,  pushing 

in  the  base  to  any  desired  point  and  noticing  the  increase 
in  altitude. 

414.  In  408  you  changed  one  quadrilateral,  A  B  C  D, 
Fig.  62,  into  four  triangles  by  using  different  guide  lines, 
D  B  and  C  A,  and  by  moving  C  and  D  first  in  one  direc- 
tion and  then  in  the  opposite  direction.  Place  these  four 
triangles  on  one  line,  ride  by  side,  and  shorten  all  the 
bases  to  one  length,  thive-fourths  of  the  line  A  B.  h 
your  work  is  accurately  done,  where  should  the  vertices 
of  the  raised  triangles  bo  ?  Why  ?  Take  pains  to  have 
this  construction  an  almost  perfect  specimen  of  workman- 
ship. 

415.  In  413  you  shortened  the  base  line  just  as  much  or 
as  little  as  you  pleased,  but  the  vertex  rose  to  a  point  that 
was  out  of  your  control.  You,  of  course,  knew  that  the 
more  you  shortened  the  base  the  higher  the  vertex  rose, 
but  you  could  not  fix  the  level  to  which  the  vertex  would 
rise,  at  least  without  further  construction  or  arithmetical 
calculation.  Can  you  raise  a  triangle  to  any  desired  alti- 
tude ?    Try  to  solve  the  problem  by  yourselves,  but,  if  you 


CONSTRUCTIONAL   GEOMETRY 


83 


is 


find  it  too  hard,  study  Fig.  64,  where  /\  A  B  C  is  the  orig- 
inal triangle  and  H  P  is  the  desired  altitude.  M  B  is  the 
guide  line,  parallel  to.  which 
you  draw  C  Q,  to  determine 
the  length  A  Q  of  the  short- 
ened base.  This  problem  is 
just  the  reverse  of  413, 
and  you  can  easily  see  that 
your  construction  is  right  by 
shortening  the  base  line  to 
A  Q,  and  by  noticing  that 
you  would  move  B  along  a 
line  symparallel  with  Q  0  to 

M.     If  Q  C  is  the  guide  to  B  M  in  one  construction,  it 
not  hard  to  see  that  M  B  becomes  the  guide  to  C  Q  in  the 
reverse  construction. 

416.  Make  three  different  triangles  and  raise  the  vertex 
to  a  level  fixed  upon  beforehand. 

417.  In  415  we  shortened  the  base  by  cutting  off  some- 
thing at  the  right  end,  moving  B  in  as  far  as  Q.  Try  to 
solve  the  same  problem  by  cutting  off  a  length  at  the  left 
end,  thus  pushing  A  in  toward  B.  Suggestion  :  Extend 
B  C  until  it  crosses  P  M  at  N,  Fig.  64,  and  draw  a  new 
guide  line,  N  A.     Compare  the 

resulting  base  line  with  A  Q  of 
Fig.  64. 

418.  Make  three  triangles, 
and  raise  the  vertices  to  any- 
desired  level  by  pushing  the 
left  end  of  the  base  toward  the 
right  end. 

Practise  changing  triangles 
to  triangles  of  greater  altitude 
until  it  is  a  matter  of  indiffer- 
ence to  you  on  which  side  of 
the  triangle  you  draw  your  construction  lines. 

419.  Can  you  now  take  the  necessary  steps  to  lower  the 


84:  ELEMENTARY  AND 

vertex  of  a  triangle  without  changing  the  area  ?  Make  a 
triangle,  ABC,  and  lower  the  vertex  to  the  level  of  P,  Fig. 
65.  Do  this  twice  :  first,  so  that  the. position  of  A  shall  not 
be  changed ;  second,  so  that  the  position  of  B  shall  not  be 
changed.     Compare  the  two  bases  obtained. 

4901  Place  two  triangles  of  different  altitudes  side  by 
side.  First,  raise  the  lower  one  to  the  level  of  the  other, 
and  add  the  two  triangles ;  second,  lower  the  taller  one  to 
the  level  of  the  shorter,  and  add  the  two.  Compare  the 
two  triangles  resulting  from  addition  in  the  two  cases  by 
changing  one  to  the  level  of  the  other. 

421.  Place  two  triangles,  one  of  which  is  a  great  deal 
taller  than  the  other,  side  by  side.  Change  them  to  a  com- 
mon level  by  raising  one  and  lowering  the  other ;  and  then 
add  the  results.  In  changing  levels  work  from  the  left 
side  of  the  left-hand  triangle,  and  from  the  right  side  of  the 
right-hand  triangle,  that  the  vertex  where  the  triangles  join 
may  not  be  changed  in  the  process. 

422.  Place  three  triangles  of  different  heights  side  by 
side  as  in  Fig.  G6. 


1 1..  «« 


Change  ^  A  B  C  and  D  F  G  to  the  level  of  A  B  D  E 
without  changing  the  positions  of  B  and  D.  Add  the  re- 
sulting triangles.  Aim  to  complete  the  problem  in  as  few 
steps  as  possible,  indicating  your  steps  by  the  kind  of  line 

drawn. 


CONSTRUCTIONAL  GEOMETRY  85 

423.  Make  a  triangle  and  a  quadrilateral  that  seem  to 
you  to  have  the  same  area.  Change  the  quadrilateral  to  an 
equivalent  triangle,  and  change  this  triangle  to  the  level  of 
the  first  triangle,  and  draw  a  triangle  representing  your 
error  in  estimation,  if  there  be  any  error. 

424.  Make  a  quadrilateral  free  hand  that  seems  equivalent 
to  the  sum  of  a  triangle  and  a  quadrilateral  that  you  have 
drawn.     Test  your  accuracy  as  in  423. 

425.  Change  a  pentagon  into  an  equivalent  triangle. 
Suggestion  :  First  mould  the  pentagon  into  a  quadrilateral, 
and  then  mould  the  quadrilateral  into  a  triangle.  The 
drawing  is  much  less  confused,  if  you  work  on  different 
sides  of  the  figure  in  taking  the  two  steps. 

426.  Find  a  triangle  equivalent  to  the  sum  of  a  pentagon 
and  a  quadrilateral. 

427.  Can  you  change  a  triangle  into  a  pentagon  ? 

428.  Figures  having  six,  seven,  eight,  or  any  number  of 
sides  can  be  changed  to  triangles  by  reducing  the  number 
of  sides  one  at  a  time,  just  as  the  pentagon  in  425  was 
changed  to  a  quadrilateral,  and  the  resulting  quadrilateral 
was  changed  to  a  triangle.  After  two  steps  the  lines  be- 
come confused,  unless  a  new  figure  is  drawn. 

All  plane  figures  that  are  bounded  by  straight  lines  are 
called  polygons.  Special  names  have  already  been  given  to 
polygons  of  three,  four,  five,  and  six  sides.    What  are  they  ? 

429.  Can  you  give  a  rule  for  changing  any  polygon  into 
an  equivalent  triangle,  that  will  hold  good  no  matter  how 
many  sides  the  polygon  has  ? 

430.  Draw  a  circumference  and  divide  it  into  twenty- 
four  equal  parts.  Can  you  do  this  without  a  protractor  ? 
Form  a  polygon  of  twenty-four  sides  by  joining  the  points 
of  division.  The  polygon  is  inscribed  in  the  circumference. 
What  is  meant  by  this  statement  ?  This  polygon  can  be 
moulded  into  a  triangle  by  a  process  much  simpler  than  the 
one  which  you  have  described  in  429  ;  for,  by  joining  the 
vertices  with  the  centre  of  the  circumference,  you  form 
twenty- four  equal  triangles.     See  404,  where  a  hexagon  was 


86  ELEMENTARY  AND 

divided  in  this  way.  Since  the  twenty-four  triangles  have 
the  same  altitude,  they  can  be  changed  into  one  triangle 
very  easily.  Describe  the  method.  How  long  a  base 
will  the  resulting  triangle  have  ?     How  tall  will  it  I 

431.  Describe  how  you  would  find  a  polygon  of  forty- 
eight  sides  inscribed  in  a  circumference  from  the  inscribed 
polygon  of  twenty-four  sides  of  430.    How  could  you  change 
this  polygon  into  a  triangle?     How  long  would  the  I 
be-' 

432.  In  the  same  way  that  you  got  an  inscribed  polygon  of 
48  sides  from  one  of  24  sides,  you  could  get  one  of  %  sides 
from  one  of  48  sides,  and  then  you  could  get  polygons  of 
r*.\  384,  70s  sides  in  succession,  continuing  the  process 
until  the  polygons  seemed  to  the  eye  to  coincide  with  the 
circle.  Each  time  the  polygon  could  be  moulded  into  a  tri- 
angle having  a  base  equal  to  tlie  perimeter  of  the  polygon 
and  an  altitude  equal  t<»  the  distance  from  the  centre  of  the 
circumference  to  one  of  the  sides  of  the  polygon. 

438.  It  is  not  difficult  to  see  from  482  that  fche circle 
itself  could  be  moulded  into  a  triangle  having  a  base 
equal  to  the  oiroumferenoe  and  an  altitude  equal  to  the 
radios.  The  difficulty  if  to  find  a  straight  line  equal  to 
the  circumference.  Make  the  following  experiments  and 
bring  your  results  into  the  class  :  Roll  a  cylinder  of  wood 
about  two  inches  in  diameter  in  a  straight  line,  making  it 
turn  ten  times.  By  marking  the  lowest  point  of  the  cylin- 
der at  the  start  and  the  |>oint  on  the  table  just  below  this 
point,  it  is  |  simple  matter  to  find  the  distance  rolled  and 
the  number  of  turns  made.  Compare  the  distance  rolled 
with  the  diameter.  Push  a  28-inch  bicycle  until  the  wheel 
has  made  ten  revolutions  Measure  the  distance  covered 
and  compare  it  with  the  diameter.  Perform  each  experi- 
ment several  times  and  record  the  results  in  a  neat  table. 

How  many  times  as  long  as  the  diameter  do  you  find  the 
circumference  to  be  on  the  average  ? 

434.  Using  the  results  of  433,  change  five  different 
circles  into  essentially  equivalent  triangles. 


CONSTRUCTIONAL   GEOMETRY 


87 


435.  Can  you  tell  why  it  is  a  much  more  difficult  prob- 
lem to  change  a  triangle  into  a  circle  ?  Suggestion  : 
Think  how  much  longer  than  the  altitude  the  base  of  the 
triangle  ought  to  be,  to  enable  you  to  call  the  altitude  of 
the  triangle  the  radius  of  the  equivalent  circle. 

436.  Draw  a  particular  triangle  that  you  can  turn  into 
a  circle  with  as  much  accuracy  as  the  results  of  433  allow. 


SECTION  XVI.     PECULIAEITIES  OF  SQUAEES. 


437.  Make  an  isosceles  right  triangle.  The  problem  is 
to  change  it  into  an  equiv- 
alent right  triangle  with  a 
leg  equal  to  any  chosen 
length,  say  P  Q,  longer 
tha^n  A  B.  One  way  of 
solving  this  problem  is  to 
raise  the  vertex  by  the 
method  of  415,  until  the 
height  A  R  is  equal  to 
P  Q.  Take  the  necessary 
steps  and  find  the  equiva- 
lent triangle  A  S  R. 

438.  A  much  more 
roundabout  way  of  solv- 
ing the  problem  of  437 —  - 
but  for  the  present  pur-  f 
poses  a  better  way — is  to 
proceed  as  follows :  move  C,  Fig.  68,  along  a  line  parallel 
to  B  A,  until  it  reaches  a  point  E  that  is  at  a  distance  P  Q 
from  A,  thus  moulding  /\ABC  into  /\ABE. 

Now  imagine  /\ABE  to  be  pivoted  at  A,  and  to  be 
turned  through  a  quarter  of  a  revolution  into  the  position  of 
/\ACR.     AA^oAABC.     Why?    Now  move  C 


Fig.  67. 


88 


ELEMENTARY  AND 


parallel  to  A  R  to  S,  thus  moulding  /\ACR  into  ^ASR. 
/\ACRoAASR-  Why?  HenceAASRoABC; 
since  /\ASR  lias  the  desired  leg  A  R  =  A  E  =  P  Q,  the 
problem  is  solved.  Compare  the  result  with  the  result  of 
the  last  article. 

439.  Using  the  method  of  438,  turn  isosceles  right  tri- 
angles into  right  triangles  with  one  leg  of  any  chosen  length 


\  *x.         // 

X                     V 

X.                  x 

\    "*•».  ' 

\                   \ 

\  -V- 

X                   \ 

X           \ 

\-'   .'   "* 

x          * 

*)x/ 

^         \     \ 

^.   \    * 

/  /N 

I 


//' 


Flo.  68. 

longer  than  the  leg  of  the  isosceles  triangle,  until  you  be- 
come perfectly  familiar  with  the  process. 

440.  What  is  the  angle  formed  by  S  C  and  S  A,  Fig.  68  ? 
Why  ?    What  kind  of  triangle  is  A  A  C  E  ? 

441.  Can  you  turn  an  isosceles  right  triangle  ABC  into 
a  right  triangle  ASR,  Fig.  68,  without  actually  drawing 
lines  B  E  and  C  R  ?  Try  three  cases  of  this  kind.  The 
lines  should  be  imagined,  though  not  drawn. 

442.  The  isosceles  right  triangle  A  B  C,  Fig.  68,  is  one 
half  of  a  square,  and  the  right  triangle  A  S  R  is  one  half  of 


CONSTRUCTIONAL   GEOMETRY 


89 


Fig.  69. 


a  rectangle.    Draw  a  figure  showing  the  square  and  rectangle 
meant.     What  is  true  of  their  areas  ?    Why  ? 

The  process  described  in  438  gives  you  the  means  of 
moulding  a  square  into  a  rectangle  having  one  side  any  de- 
sired length  longer  than  the  side  of 
the  square.     Do  you  see  how  ? 

443.  Turn  four  squares  into  rectan- 
gles. 

444.  Turn  one  square  into  four  dif- 
ferent rectangles. 

445.  In  437  and  438  we  have  seen 
how  to  turn  an  isosceles  right  triangle 
into  a  right  triangle  with  unequal  legs, 
in  two  different  ways.  Can  a  right 
triangle  with  unequal  legs  be  changed 
to  an  isosceles  right  triangle  ?  Try 
first  to  reverse  the  method  of  437,  and 
to  change  a  right  triangle  A  S  E,  Fig.  69,  to  an  isosceles 
right  triangle  A  B  C  by  lowering  the  vertex  E  to  the  proper 
level  C,  so  that  the  extended  base  A  B 
shall  be  equal  to  the  reduced  height 
A  C.  How  will  you  take  the  first  step  ? 
Why  is  it  difficult  to  draw  a  guide  line, 
parallel  to  which  E  may  be  moved  ? 

446.  In  the  next  place,  try  to  reverse 
the  method  of  438.  To  help  the  imag- 
ination, place  triangle  A  S  E  (Fig.  70) 
in  a  position  something  like  that  of  tri- 
angle A  S  E  in  Fig.  68. 

The  problem  still  is  to  find  C,  to  which 
to  move  the  vertex  S.  We  know  that  C 
is  somewhere  on  a  line  drawn  parallel  to 
E  A  from  S,  but  we  do  not  know  exactly 
where.  It  is  through  the  triangle  corre- 
sponding to  A  A  C  E  (Fig.  68)  that  we 
shall  be  able  to  find  C.  In  440  you  learned  that  /\ACE 
is  a  right  triangle.     How  can  you  find  the  point  corre- 


Fig.  70. 


90  ELEMENTARY  AND 

sponding  to  E  ?  Review  272,  and,  with  the  aid  of  the 
principle  therein  stated,  find  point  C.  A  C  will  be  the  side 
of  the  isosceles  triangle  desired.  Draw  A  13  at  right  angles 
to  A  C  and  equal  to  it.     Complete  triangle  ABC. 

447.  Change  six  different  right  triangles  into  isosceles 
right  triangles  by  the  method  of  44G.  Vary  the  position 
of  the  triangles,  that  you  may  not  depend  upon  having  your 
figure  placed  as  in  Fig.  08. 

44*.  Change  a  rectangle  into  a  square,  following  the 
method  suggested  by  44M.     Hepeat  three  times. 

449.  Change  a  triangle  into  a  square.  Suggestion  :  First 
change  the  triangle  into  a  rectangle. 

450.  Change  a  pentagon  into  a  square.  Make  the  steps 
in  their  proper  order  perfectly  clear. 

•l"'l.  Add  a  triangle  and  a  quadrilateral,  and  change  the 
resulting  figure  into  a  square. 

452.  Put  three  equal  squares  side  by  side,  so  as  to  form  a 
rectangle.  Mould  the  rectangle  thus  formed  into  a  square 
form. 

Find  a  square  three  times  as  large  as  a  given  square. 
Find  a  square  five  times  as  large  as  a  given  square. 

454.  Cut  a  square  into  five  equal  rectangular  strips,  find- 
ing J  of  one  side  with  your  dividing  tool.  Change  one  of 
these  strips  into  a  square  form.  How  is  this  new  square  re- 
lated to  the  original  square  ? 

455.  Find  i  Bqnare  \  of  a  given  square.     See  454. 
466.    Find  ■  square  ^  of  a  given  square.     See  454. 

457.  Make  a  pentagon.  Can  you  find  a  square  that  shall 
be  j  as  large  as  the  pentagon  ? 

458.  In  changing  a  square  into  a  rectangle,  438,  yon 
found  that  a  gnat  many  rectangles  could  be  formed  equiva- 
lent to  the  square.  The  shape  of  the  rectangle  depends 
upon  the  distance  which  the  vertex  of  the  square  is  moved 
along  C  E  (see  Fig.  68).  In  changing  a  rectangle  into  a 
square,  how  many  results  can  you  get  ? 

459.  Make  a  square,  A  B  C  D,  Fig.  71.  Change  the 
square  into  a  rectangle  eight  times  by  moving  each  vertex 


CONSTRUCTIONAL  GEOMETRY 


91 


in  the  two  directions  indicated  by  the  arrows  at  the  vertex 
(Pig.  71). 

Move  each  vertex  the  same  distance,  and  compare  the  re- 
sulting rectangles. 

Note. — The  object  of  this  ex- 
ercise is  to  give  you  practice  in 
changing  squares  into  rectangles 
with  the  squares  in  different 
positions.  If  you  find  any  one 
position  hard  for  you,  practise 
that  case  until  it  is  as  simple  as 
any  of  the  others. 

460.  Draw  a  square  ABCD 
(Fig.  72),  and  change  it,  by 
438,  to  rectangle  K  H  A  F.  The 
fact  that  the  square  is  equivalent  to  the  rectangle  may  be 

"  OAF  x  AH. 
x  ES 


JK                                   * 

V- 

D                      C 
A                           R 

cl                  +' 

Fig.  71. 


]  S  K  F  E. 


expressed  as  follows  :  A  D 

Show  that  ED2=G=EP 

Suggestion  :  Imagine  that  D 
has  been  moved  to  A,  parallel 
toNE. 

461.  What  kind  of  quadri- 
lateral is  S  H  A  E,  which  is 
made  up  of  the  two  rectangles 
S  K  F  E  and  K  H  A  F  ? 

Show  that  AE2  O  AD2  + 
D~E2- 

462.  Does  the  truth  stated  in 
461  apply  to  all  right  trian- 
gles ? 

463.  The  principle  of  460  and 
461  stated  in  words  is  as  fol- 
lows :  The  square  formed  on  the  hypotenuse  of  a  right  tri- 
angle is  equivalent  to  the  sum  of  the  squares  formed  on 
its  sides. 

This  principle  is  called  the  Pythagorean  principle,  be- 


92  ELEMENTARY  AND 

cause  the  discovery  of  it  is  attributed  to  Pythagoras,  a  famous 
Greek  mathematician,  who  lived  about  600  B.C.,  and  who 
did  much  to  encourage  the  study  of  Geometry.  How 
Pythagoras  made  his  discovery  is  not  known.  The 
method  of  showing  its  truth  here  given  is  essentially  that 
devised  by  Euclid,  who  lived  three  hundred  years  later  than 
Pythagoras,  and  who  was  the  most  famous  of  the  ancient 
Geometers. 

404.  The  Pythagorean  principle  makes  the  addition  of 
squares  even  simpler  than  the  addition  of  triangles  ;  for  it 
is  necessary  merely  to  place  the  squares  to  be  added  on  the 
legs  of  an  imaginary  right  triangle,  as  in  Fig.  72.  The 
square  on  the  hypotenuse  is  the  square  desired. 

465.  Add  two  squares.  Repeat  the  process  four  times, 
using  different  squares. 

466.  Can  you  form  two  unequal  squares  on  one  line,  A  B, 
as  a  side  ?  If  A  B  is  a  given  line,  can  you  form  AB2  with- 
out any  further  description  ? 

What  is  meant  by  the  statement  that  a  square  is  deter- 
mined by  one  of  its  sides  ? 

467.  Can  von  find  the  side  of  a  square  equivalent  to 
MN*  +  §T,  where  M  N  and  S  T  are  given  lines,  without 
constructing  any  square  ?  Describe  your  process  carefully. 
It  is  better,  in  the  majority  of  cases,  not  to  draw  the  squares 
actually,  but  to  use  the  sides  of  the  squares. 

408.  Given  two  lines,  a  and  b,  find  «*  +  Ir  in  n  square  form. 

Find  a  square  equivalent  to  a*  +  a7  or  2  a1 ;  to  «*  +  2  «* 
or  8  n2.  Find  a  square  equivalent  to  3  «*  by  453,  and  com- 
pare the  result  with  that  just  obtained.  Compare  your 
method  <>f  getting  the  side  of  2a*  here  with  the  method 
suggested  in  315. 

469.  Given  three  lines  a,  b,  c.  Can  you  find  a  square 
equivalent  to  a*  +  #*  +  c*?  Find  a*  +  a*  +  a*  by  this 
method.  Find  3a8  by  the  methods  of  453  and  468,  and 
compare  the  results  with  that  obtained  here. 

470.  Change  two  triangles  into  squares,  and  add  the  re- 
sulting squares.     Add  the  two  triangles  without  changing 


CONSTRUCTIONAL  GEOMETRY  93 

them  to  squares  ;  change  the  resulting  triangle  to  a  square, 
and  compare  the  results  obtained  by  the  two  methods. 

471.  Make  a  square  on  a  given  line,  a.  Make  a  square 
on  a  line  equal  to  2a.  Compare  the  two  squares,  a2  and 
(2a)2.  How  much  larger  than  the  first  square  is  the 
second  ? 

472.  Make  a  square  on  the  line  da  as  a  base,  and  com- 
pare the  resulting  square  with  a2. 

473.  What  is  the  side  of  a  square  equivalent  to  16  a2  ?  to 
25  a2  ?  to  100  a2  ? 

474.  Make  a  square  on  the  line  a.  Draw  a  line  M  N" 
equal  to  3a  and,  at  right  angles  to  M  N,  draw  N  P  equal 
to  a.  Join  M  P,  and  call  it  I.  How  many  times  as  big 
as  a2  is  b2  ? 

475.  Can  you  devise  a  simple  method  of  finding  5  a2  from 
the  experiment  in  474  ?  Find  a  square  five  times  as  large 
as  a2  by  the  method  of  453,  and  compare  the  result  with 
that  obtained  by  your  new  method. 

476.  Can  you  draw  a  right  triangle  by  first  drawing  the 
hypotenuse  ?  Eecall  what  you  know  about  the  middle 
point  of  the  hypotenuse. 

477.  Make  a  line  A  B,  and  draw  a  semi-circumference, 
of  which  A  B  is  the  diameter.  Join  any  point,  C,  of  this 
semi-circumference  with  A  and  B.  Can  you  tell  the  size 
of  /  B  C  A  ?     Give  a  reason  for  your  answer. 

478.  Make  a  square,  a2,  of  the  same  size  as  that  used  in 
474  and  475.  With  a  line  A  B 
equal  to  3a  as  a  diameter,  de- 
scribe a  semi-circumference.  See 
Fig.  73.  Make  a  line  AC, 
called  a  chord  of  the  circum- 
ference, equal  to  a.  Draw  the 
chord  C  B.  Fig.  T3. 

Draw  squares  on  A  B,  A  C,  and  B  C. 

How  much  larger  than  A  C2  is  A  B2  ? 

How  much  larger  than  AC2  is  CTB2  ?    Why  ? 

479.  Draw  a  new  semi-circumference  of  the  same  size  as 


94  ELEMENT WHY  AND 

that  of  478,  and  this  time  draw  a  chord  A  D  equal  to  2a. 
Join  D  B,  draw  the  squares  as  before,  and  compare  all  the 
squares  with  a2. 

480.  Draw  two  squares,  a2  and  #*.  Can  you  find  a  square 
equivalent  to  the  difference  between  a2  and  #*  ?  Describe 
the  steps  carefully. 

481.  Cut  from  a  square  piece  of  paper,  at  any  place,  a 
smaller  square.  If  the  remaining  piece  of  paper  could  be 
moulded  into  a  square  form,  how  long  would  the  sides  of 
the  square  be  ?  Show,  by  a  neat  construction,  how  you 
would  obtain  the  desired  length. 

482.  Make  three  pairs  of  squares  and  subtract  the  smaller 
square  of  each  pair  from  the  larger,  obtaining  your  result 
in  the  form  of  a  square. 

483.  It  is  a  simple  matter,  as  you  have  seen,  to  find  a 
square  equal  to  4a2,  9a2,  10a2,  25a2,  etc.,  when  a2  isght  n. 
Can  you,  by  simple  addition  or  subtraction  of  these  easily 
obtained  squares,  find  squares  equal  to  7a2,  13a2,  12a2,  20a2, 
3a2? 

Whenever  you  see  two  combinations  that  will  give  the 
same  result,  it  is  well  to  try  both  and  to  compare  the  an- 
swers. After  the  first  two  or  three  problems,  it  is  not 
necessary  to  draw  the  actual  squares.  Y<>u  will  find  that 
you  need  to  deal  with  the  sides  only  ;  for  a  square  is  easily 
formed  in  imagination,  if  its  side  is  clearly  seen. 

484.  Not  receiving  sufficient  Light  from  a  certain  square 
window,  a  man  deeided  to  make  an  opening  twice  as  large 
without  changing  its  shape;  he,  accordingly,  made  his 
window  twice  as  long  and  twice  as  wide  as  before.  How 
much  more  light  was  let  in  ?  What  should  he  have  done 
to  find  the  proper  length  for  his  window  ? 

485.  Two  farmers  have  square  fields  to  fence  ;  if  the  field 
of  the  first  farmer  is  four  times  as  large  as  that  of  the 
second,  how  much  more  fencing  will  he  need  ?  Draw 
plans  of  the  fields  before  making  a  final  answer  ?  Can  you 
account  for  the  difference  between  the  relative  cost  of  fenc- 
ing and  the  relative  size  of  the  fields  ? 


CONSTRUCTIONAL  GEOMETRY  95 

486.  On  a  piece  of  cardboard  draw  a  diagram,  showing 
how  you  would  cut  the  cardboard  to  make  it  fold  into  a 
hollow  cube  with  one-inch  edges.  If  you  were  to  make  a 
hollow  cube  with  two-inch  edges,  how  much  more  surface 
would  your  diagram  cover  ?  How  much  more  sand  would 
your  second  cube  hold  than  the  first  would  hold  ?  How 
much  cardboard  would  you  save  in  making  one  large  cube 
like  the  second  one,  in  place  of  making  enough  small 
cubes  like  the  first  to  hold  an  equal  amount  with  the 
second  ? 

487.  If  you  want  to  make  a  cubical  box  that  will  have 
an  outside  surface  just  twice  as  large  as  that  of  the  first 
cubical  box  of  486,  how  will  you  proceed  to  draw  the  dia- 
gram by  which  to  cut  out  the  box  ? 

488.  Can  you  draw  a  diagram  that  will  fold  into  a  cube 
with  a  surface  three  times  as  large  as  that  of  the  first  cube 
of  486  ? 

489.  You  cannot  become  too  familiar  with  the  peculi- 
arities of  squares  that  have  been  dwelt  upon  in  this  sec- 
tion. You  will  find  later  that  a  great  many  other  figures 
behave  like  squares  ;  but  this  will  do  you  little  good,  un- 
less you  know  well  the  peculiar  characteristics  of  squares, 
which  enable  you  to  add  them,  to  subtract  them,  to  show 
what  effect  the  change  in  a  side  will  produce  upon  the  sur- 
face, and  to  show  how  a  change  in  the  size  of  the  square 
affects  the  side.  You  ought  also  to  be  expert  in  moulding 
any  polygon  into  a  square  form. 

490.  Cut  from  a  piece  of  paper  a  square  3  in.  by  3  in. 
Can  you,  by  adding  strips  to  two  sides  only,  keep  the 
paper  in  the  form  of  a  square  ? 

If  you  add  strips  an  inch  wide,  how  long  must  the  strips 
be  to  keep  the  square  perfect  ?  How  many  square  inches 
in  the  strips  added  ? 

491.  Cut  a  strip  J  in.  wide  of  the  right  length  to  make 
a  perfect  square  when  added  to  two  sides  of  the  3  in. 
square  of  490.     How  many  square  inches  in  the  strip  ? 

If  the  strip  is  2  in.  wide,  how  long  should  it  be  ? 


96  ELEMENTARY  AND 

492.  Imagine  a  square  7  ft.  by  7  ft.  How  long  a  strip 
will  be  needed  to  enlarge  the  square  to  a  square  9  ft.  by  9 
ft.,  if  the  strip  is  added  on  two  sides  ?  How  many  square 
feet  in  the  strip  ? 

493.  Imagine  a  square  80  ft.  by  80  ft.  How  long  a  strip 
will  be  needed  to  enlarge  this  to  a  square  87  ft.  by  87  ft.  ? 
How  many  square  feet  in  the  strip  ? 

494.  Calculate  the  lengths  and  areas  of  the  strips  needed 
in  each  of  the  following  cases  : — square  70  ft.  by  70  ft., 
strip  9  ft.  wide  ;  square  130  ft.  by  130  ft.,  strip  5  ft.  wide  ; 
square  90  ft.  by  90  ft.,  strip  6  ft.  wide  ;  square  700  ft.  by 
700  ft.,  strip  20  ft.  wide. 

495.  Can  you  state  the  law  by  which  you  can  find  the 
length  of  the  strip  from  the  length  of  the  square  and  the 
width  of  the  strip  ? 

496.  Make  a  square  3  in.  by  3  in.  If  you  have  sixteen 
square  inches  of  paper  which  can  be  cut  into  rectangular 
strips,  how  wide  will  you  make  your  strips,  that  you  may 
have  a  perfect  square  when  you  shall  have  added  the  strips 
to  the  3  in.  square  ?  The  strips  in  this  and  the  follow- 
ing problems  are  to  be  added  on  only  two  sides  of  the 
square. 

Solve  the  problem  of  496,  if  the  original  square  is 
9  ft.  by  9  ft.  and  the  material  for  addition  has  a  surface  of 
63  sq.  ft. 

498.  By  adding  a  strip  that  contained  889  sq.  ft.  to  a 
square  60  ft.  by  60  ft,  a  square  was  obtained  ;  how  wi<l< 
and  how  long  was  the  strip  ? 

499.  Calculate  the  lengtli  and  width  of  the  strips  in  the 
following  cases  : — square  80  ft.  by  80  ft.,  area  of  strip  656 
sq.  ft.  ;  square  50  ft.  by  50  ft.,  area  of  strip  981  sq.  ft.  ; 
square  170  ft.  by  170  ft.,  area  of  strip  2076  sq.  ft. 

500.  Can  you  make  a  rule  for  finding  the  length  and 
width  of  a  strip  needed  to  make  a  perfect  square,  when 
added  to  a  known  square,  from  the  side  of  the  known 
square  and  the  area  of  the  strip  ? 

501.  Arrange  784  sq.  ft.  in  the  form  of  a  perfect  square 


CONS  TE  UGTIONAL  GEOMETR  Y 


97 


by  first  making  as  large  a  square  as  you  can,  using  only 
round  numbers  like  10,  20,  50,  100,  300,  etc.,  for  the  side 
of  the  square,  and  by  then  adding  the  material  left  in  the 
form  of  a  strip  according  to  the  examples  given  in  articles 
496-499,  and  according  to  the  rule  in  500. 

The  following  model  for  the  arrangement  of  the  work  is 
suggested.  In  the  model  the  problem  is  to  arrange  4489 
sq.  ft.  in  a  perfect  square. 


Dimensions 
of  strip. 


0 
127 

7 


Material. 
4489 
3600 


889 


Side  of  square. 
60  (1st  guess) 

7  (width  of  strip) 


67  answer. 


Explanation.  —  A  square  70 
ft.  by  70  ft.  would  require  4900 
sq.  ft.  of  surface  ;  therefore,  in 
round  numbers,  a  square  60  ft.  Fig.  u. 

by  60  ft.  is  the  largest  possible 

square.  The  material  left  after  forming  this  square  (Fig. 
74),  is  889  sq.  ft.  ;  the  length  of  the  strip  to  be  added  must 
be  at  least  2  x  60  ft.  or  120  ft.  ;  the  width  is  roughly  889 
-T-  120  or  7  ;  since  the  strip  must  be  as  much  longer  than 
120  ft.  as  it  is  wide,  7  ft.  is  added  to  120  ft.  to  obtain  the 
true  length  of  the  strip.  The  surface  covered  by  the  strip 
is  127  ft.  x  7  ft.  or  889  sq.  ft.  All  the  material  is  thus 
used,  and  a  perfect  square  is  formed.  In  writing  the  first 
estimate,  120  ft.,  of  the  length  of  the  strip,  the  cipher  was 
written  a  little  above  the  line,  to  leave  room  for  the  addi- 
tion of  the  width  to  the  length  as  soon  as  the  width  could 
be  guessed.  It  is  well  to  draw  a  figure  like  Fig.  74  to  aid 
the  imagination,  until  the  process  is  perfectly  familiar  to 
you,  so  that  no  aid  is  needed. 

502.  Arrange  the  following  surfaces  in  square  forms,  fol- 
lowing the  model  given  in  501  :  1156  sq.  ft.  ;  6889  sq.  ft.  ; 
7 


98 


ELEMENTARY  AND 


5184  sq.  ft.  ;  2401  sq.  in.  ;  1G9  sq.  yds.  ;  729  sq.  ft.  : 
sq.  in.     In  doing  the  last  problem  follow  this  model  : 


Ditnetiftions 
of  strip. 


Material .  Side  of  sq  uare 

418609    600  (1st  guess) 
B0OOOO 


width  of  1st 
strip 


640i8lde  of  M 
\       square 


width  of  2d 
strip 

answer. 


7ft                 foJi-Jht 

uo* 

Log* 

«M 

Fio.  75. 


In  getting  the  width  of  the 
first  strip,  do  not  try  to  guess  anything  except  round  num- 
bers like  30,  40,  50,  etc.  ;  and  do  not  try  to  add  all  of  the 
material  in  one  strip.  The  explanation  is  the  same  as  that 
given  in  the  previous  article  ;  two  steps  instead  of  one  are 
necessary. 

503.   Arrange  the  following  surfaces  in  a  square  form,  ob- 
taining the  length  of  the  side  of  the  square  in  each  case: 
<j.  ft.  ;  146161  sq.  ft. ;  98686  sq.  ft. ;  00516  sq.  ft.  ; 
120409  sq.  ft. 

I  n  arranging  the  following  surfaces  in  a  square  form, 
you  will  find  that  some  material  will  be  leftover.  Calcu- 
late the  side  of  the  square  and  the  amount  left  over  in  each 
case  :— 536  sq.  ft.  :  4376  sq.  ft.  ;  59785  sq.  ft. 

505.  If,  after  adding  one  or  two  strips  to  a  square,  there 
is  not  enough  material  left  to  make  a  strip  1  ft.  wide  or  1 
in.  wide,  according  to  the  unit  used,  it  is  possible  to  add 
strips  that  are  so  many  tenths  of  a  foot,  or  tenths  of  an  inch 
in  width  :  and  then  to  add  other  strips  whose  width  is 
measured  in  hundredths  of  a  foot  or  hundredths  of  an  inch, 
and  so  on  until  the  material  left  is  not  worth  noticing.  See 
the  following  example,  where  281  sq.  ft.  are  to  be  arranged 
in  the  form  of  a  square  : 


CONSTRUCTIONAL   GEOMETRY 


99 


Dimensionsof  strip. 

0 

26 

Material. 

281. 
100. 

10 
6 

Side  of  square, 
first  guess 

181. 
156. 

width  of  1st  strip 

6 

16. 

.7 

side  of  2d  square 

33.7 

.7 

25.00 
22.89 

width  of  2d  strip 

33.46 

.06 

2.1100 
2.0076 

16.7 
.06 

side  of  3d  square 
width  of  3d  strip 

33.523 

.003 

.102400 
.100569 

16.76 
.003 

side  of  4th  square 
width  of  4th  strip 

.001831 

16.763 

answer. 

281  sq.  ft.,  therefore,  when  moulded  into  the  form  of  a 
square,  will  make  a  square  whose  side  is  16.763  ft.  ;  the 
part  unused,  a  little  under  two-thousandths  of  a  square 
foot,  is  not  worthy  of  notice  in  ordinary  cases.  Multiply 
16.763  by  16.763,  and  see  how  much  the  product  lacks  of 
amounting  to  281. 

506.  Arrange  in  the  form  of  a  square  the  following  sur- 
faces ;  calculate  the  side  to  the  thousandth  of  your  unit : — 
69  sq.  ft.  ;  351  sq.  ft.  ;  48  sq.  ft.  ;  10  sq.  in.  ;  5  sq.  in.  ;  3 
sq.  in.  ;  2  sq.  in. 

507.  Make  a  square  decimeter.  Draw  the  diagonal  and 
measure  it  carefully.  Arrange  in  the  form  of  a  square  two 
square  decimeters,  calculating  the  length  of  the  side  as  in 
506.     Compare  your  two  answers. 

508.  The  side  or  base  of  a  square  is  sometimes  called  the 
root  of  the  square.  The  square  root  of  a  surface  is  one 
of  the  two  equal  lines  that,  multiplied  together,  will  pro- 
duce the  surface  ;  in  other  words,  it  is  the  side  of  the  square 
into  which  the  surface  may  be  moulded.  Thus  the  square 
root  of  4489  sq.  ft.  in  501  was  found  to  be  67  ft.  This 
fact  would  be  written  in  sign  language  as  follows : — 
V  4489  sq.  ft.  =  67  ft.  Notice  the  sign  for  square  root. 
You  have  already  found  the  square  root  of  1156  sq.  ft., 


100  ELEMENTARY  AND 

2401  sq.   in.,   1G9  sq.  yds.,  in  502  ;  write  the  results  ob- 
tained, making  use  of  the  proper  signs. 

In  arithmetic  the  square  root  of  a  number  is  one  of  the 
two  equal  factors  that,  multiplied  together,  will  prodm-c 
the  number.  The  process  of  getting  the  square  root  of  a 
number  is  the  same  as  that  of  getting  the  square  root  of 
a  surface  explained  in  articles  501-505. 

509.  In  507  you  drew  a  line  equal  to  V  2  sq.  dm. 
without  any  arithmetical  work.  Can  you  think  of  a 
way  to  draw  a  line  equal  to  V  5  sq.  in.  ?  Compare  your 
answer  with  the  answer  found  in  50G  by  the  former  method. 

510.  Draw     lines    equal    to    V  2  sq.  in.,  V  10  sq.  in., 

V  3  sq.  in.  ;   explain  your  steps ;   measure  the  resulting 
lines  ;  and  compare  the  answers  with  those  obtained  in  500. 

511.  Can  you  draw  a  line  that  shall  represent  the 
V|sq.  in.;  V  |  sq.  in.  ;  V  \  sq.  in.  (See  454.)  In  actu- 
ally doing  this  work,  it  is  better  to  draw  to  scale,  taking 
a  lino  much  longer  than  an  inch  to  represent  the  side  of 
the  square  inch. 

512.  On  a  line  mark   <»1T   lengths  equal    to    \     L  tq.  dm., 

V  2  sq.  dm.,  V  3  sq.  dm.,  etc.,  up  to  V  10  sq.  dm. 

513.  On  a  line  mark  off  lengths  equal  to  V  J  sq.  dm., 

V  I  sq.  dm.,  V  }  sq.  dm.,  etc.,  up  to  V  <fo  sq.  dm. 

514.  With  the  scales  you  have  just  made  in  512  and  513, 
ami  with  the  aid  of  what  you  know  about  adding  and  sub- 
tracting squares,  you  can  find  the  square  root  of  a  great 
many  surfaces  with  great  rapidity.  Use  the  scales  in  find- 
ing the  following  square  roots  :  v  17  sq.  dm.;  V 12  sq.  dm.; 

V  24  sq.  dm. ;  V  5}  sq.  dm. ;  V  2J  sq.  dm.;  V T\ sq.  dm. 

515.  Iu  articles  501-606  you  saw  how  to  find  the  side  of 
a  square  from  its  surface.  The  ability  to  do  this  adds 
greatly  to  the  practical  value  of  the  Pythagorean  principle 
that  the  square  on  the  hypotenuse  of  a  right  triangle  con- 
tains as  much  surface  as  the  squares  on  the  two  legs. 
If  the  legs  of  a  right  triangle  are  7  in.  and  3  in.,  the 
square  on  the  hypotenuse  must  contain  49  sq.  in.  +  9  sq. 


CONSTRUCTIONAL   GEOMETRY  101 

in.  or  58  sq.  in.     Show  that  the  length  of  the  hypotenuse 
is  7.61  in. 

516.  Measure  the  length  and  width  of  the  bottom  of  a 
chalk  box,  and  calculate  the  length  of  the  diagonal  across 
the  bottom. 

517.  Measure  the  length  and  width  of  the  school-room, 
and  calculate  how  far  apart  the  diagonally  opposite  corners 
are. 

518.  A  table  is  8  dm.  long  and  6.6  dm.  wide  ;  how  long 
a  straight  stick  could  be  placed  on  the  table  so  that  no  point 
of  the  stick  would  be  beyond  the  edge  of  the  table  ? 

519.  A  drawer  is  8  dm.  long,  3.5  dm.  wide,  and  1  dm. 
deep  ;  what  is  the  longest  straight  stick  that  can  be  placed 
wholly  within  the  drawer  ? 

520.  A  ladder  13  ft.  long  was  placed  against  a  tree,  with 
its  foot  5  ft.  from  the  base  of  the  tree  ;  how  far  up  the  tree 
did  the  ladder  reach  ? 

521.  A  boy  walked  round  two  sides  of  a  rectangular  field, 
11  yds.  by  60  yds. ;  how  much  would  he  have  saved,  if  he 
had  gone  diagonally  across  the  field  ? 

522.  In  raising  a  pole  65  dm.  long  a  man  rested  the  up- 
per end  against  a  house  ;  he  found  the  lower  end  to  be  33 
dm.  from  the  house  ;  the  next  time  that  he  rested  the  pole 
against  the  house  he  found  that  its  end  was  16  dm.  from 
the  house  ;  how  many  decimeters  had  he  raised  the  upper 
end  the  second  time  ?  Make  an  estimate  of  the  amount ; 
then  calculate  it ;  and  compare  the  results. 

523.  An  isosceles  triangle  has  a  base  24  in.,  and  legs  37 
in.  long  ;  how  long  is  the  altitude  of  the  triangle  ?  How 
many  square  inches  in  its  surface  ? 

524.  An  equilateral  triangle  has  sides  12  in.  long.  Can 
you  calculate  its  altitude  and  its  area  ? 

525.  Make  a  right  angled  triangle  with  one  leg  3  in.  and 
the  other  leg  4  in.     Calculate  the  hypotenuse. 

526.  The  triangle  that  you  have  just  made  is  a  most  use- 
ful one  in  practical  work.  If  you  wish  to  nail  two  straight 
sticks  together  so  that  their  edges  shall  be  at  right  angles 


102  ELEMENTARY  AND 

to  each  other,  you  can  mark  the  points  3  in.  or  3  ft.  along 
one  stick  and  4  in.  or  4  ft.  along  the  other,  and  then  nail 
a  stick  across  the  two  so  that  there  shall  he  a  length  of  5  in. 
or  5  ft.  between  the  points  marked.  See  whether  the  edges 
of  a  box  are  really  at  right  angles  to  each  other  by  this  test. 
Show  how  you  could  set  a  corner  post  at  right  angles  to  a 
platform  by  the  aid  of  the  3,  4,  5,  right  triangle. 

527.  A  pyramid  is  formed  by  four  equilateral  triangles 
resting  on  t  he  sides  of  a  square  base  G  in.  by  0  in.  Calcu- 
late the  entire  surface  of  the  pyramid. 

528.  A  rhombus  has  sides  13  in.  long  and  one  diagonal 
24  in.  long ;  how  long  is  the  other  diagonal,  and  what  is 
the  area  of  the  rhombus  ? 

529.  Divide  a  circumference  whose  radius  is  6  in.  into 
arcs  of  60  °  each  ;  join  the  ends  of  these  arcs,  forming  a 
regular  hexagon.  Divide  the  hexagon  into  six  triangles 
from  the  centre,  and  calculate  the  area  of  the  hexagon. 

530.  The  end  of  a  house  is  20  ft.  by  20  ft.  to  the  eaves; 
the  ridge  pole  is  30  ft.  above  theground.  Find  the  length  on 
the  roof  from  ridge  pole  to  eaves,  and  the  number  of  square 
feet  in  the  gable  end.     Can  you  tell  the  piteh  of  the  rouf  } 

581.  A  diamond-shaped  window  pane  has  one  edge  8  in. 
long  and  one  angle  60°.  Can  you  calculate  the  Dumber 
of  square  inches  of  glass  in  the  pane  ?  Would  more  or  less 
light  come  in,  if  the  pane  were  in  the  shape  of  a  square  with 
one  side  8  in.  } 

An  equilateral  triangle  has  sides  18  in.  long.  Draw  two 
of  its  altitudes,  ami  calculate  how  far  from  the  vertices  of 
the  triangle  they  cross  each  other. 

Divide  a  circumference  whose  radius  is  10  in.  into  arcs 
of  90°  each  :  join  the  ends  of  the  arcs,  forming  a  square. 
Calculate  the  sides  and  the  area  of  this  square. 

A  circumference  is  drawn  through  the  corners  of  a 
square  whose  sides  are  8  in.  long.  Can  you  calculate  the 
radius  of  the  circuinferen- 

The  diagonals  of  a  rhombus  are  IS  ft.  and  20  ft.  respec- 
tively.     Calculate  the  sides  of  the  rhombus. 


CONSTRUCTIONAL   GEOMETRY 


103 


532.  Models  of  solids  having  four,  eight,  or  twenty  faces 
in  the  shape  of  equilateral  triangles,  may  be  made  from  card- 
board by  drawing  the  patterns  of  Fig.  76,  and  by  cutting 
the  cardboard  half  through  along  the  lines  that  separate 
the  triangles.  Bend  the  cardboard  until  the  edges  come 
together,  and  glue  the  edges.  Calculate  the  total  surface 
of  each  solid,  if  one  side  of  the  equilateral  triangle  is  1  in. 


Fig.  76. 


104  ELEMENTARY  AND 


SECTION  XVII.     AREAS    OF  SIMILAR    FIGUBE& 


Pio.  T7. 

533.  With  your  dividing  tool,  divide  one  side  A  C  of  a 
scalene  triangle,  ABC  (Fig.  77),  into  seven  equal  parte, 
and  draw  lines  parallel  to  the  base  through  the  points  of 
division.  Draw  D  S  parallel  to  C  B,  and  draw  E  S  the 
diagonal  of  the  parallelogram  D  E  II  S.  Show  that  the 
trapezoid  V  II  K  I)  is  made  up  of  three  triangles  equal  to 
A  1>  B  (\  What  part  of  A  C  F  II  is  [\  D  C  E  ?  Bow 
much  larger  is  A  ^  H  C  than  A  D  E  C  ?  How  much 
longer  is  F  1 1  than  D  E  ?  F  C  than  DC?  II  C  than  EC? 
Into  how  many  triangles  like  DEC  can  you  divide  the 
noond  trapezoid  K  IS  II  FF  Give  the  number  of  triangles 
likeDE  0  in  the  third  trapezoid  MNRK;  in  the  fourth 
trapezoid  UVNM:  in  the  fifth  trapezoid  PQVU;  in 
the  sixth  trapezoid  A  B  Q  P.  How  much  bigger  than 
ADECisAABC?  How  much  longer  are  the  sides  of 
A  A  B  C  than  the  corresponding  sides  of  A  D  E  C  ?  Com- 
pare the  sides  of  A  C  K  R  with  the  corresponding  si 
A  C  D  E  ;  also  compare  the  surfaces  of  the  two  triangles. 


CONSTRUCTIONAL   GEOMETRY  105 

Compare  the  sides  of  /\  C  M  N  with  the  corresponding  sides 
of  /\  C  D  E  ;  also  compare  the  surfaces  of  the  two  triangles. 
Compare  the  surfaces  and  the  corresponding  lines  of 
^  C  U  Vand  C  F  H  ;  of  ^  C  P  Qand  C  M  N  ;  of  ^  C  A  B 
and  CUV. 

534.  Triangles  like  those  studied  in  533,  where  the  cor- 
responding angles  are  equal,  and  where  the  corresponding 
sides  have  the  same  relation  to  each  other,  are  called  simi- 
lar triangles.  The  investigations  suggested  in  533  point  to 
the  law  that  similar  triangles  have  the  same  peculiarities 
as  squares  in  the  matter  of  their  surfaces.  Thus  you  have 
learned  that,  to  quadruple  a  given  square,  you  make  a 
square  with  its  sides  twice  as  long  as  those  of  the  given 
square.  In  the  same  way,  to  quadruple  a  given  triangle, 
you  make  another  triangle  similar  to  it  in  shape,  with  the 
sides  respectively  double  those  of  the  given  triangle.  Com- 
pare the  triangle  in  Fig.  77,  that  is  nine  times  /\CDE 
with  /\CDE,  and  see  if  its  sides  have  the  relation  to 
those  of  /\  C  D  E  that  you  would  expect  from  your  knowl- 
edge of  squares.  Do  the  same  with  the  triangle  that  is 
twenty-five  times  as  large  as  /\  C  D  E  ;  with  the  triangle 
that  is  thirty-six  times  as  large  as  /\  C  D  E. 

535.  In  Fig.  77,  one  of  the  triangles  is  two  and  one- 
quarter  times  as  large  as  /\  C  F  H.  Find  this  triangle 
and  record  the  result  in  this  way  : — /\  ....  o  fA^FH. 
Do  the  sides  of  the  triangle  found  bear  the  relation  to 
those  of  /\  C  F  H  that  you  would  expect  from  your  knowl- 
edge of  squares  ? 

536.  The  sides  of  /\  C  U  V  are  f  of  the  corresponding 
sides  of  /\  C  F  H.  If  similar  triangles  resemble  squares  in 
the  matter  of  areas,  how  much  larger  than  /\CFH  should 
/\  C  U  V  be  ?  Test  your  answer  by  counting  the  number 
of  triangles  like  /\  C  D  E  in  each  of  the  larger  triangles 
concerned. 

537.  All  of  the  experiments  made  from  533  to  536  con- 
firm the  statement  that  similar  triangles  are  like  squares 
in  the   way  in  which  the    areas  change   as     the   sides 


106  ELEMENTARY  AND 

change.      To  give  a  complete  proof  of  this   statement 
would  carry  us  too  deeply  into  the  principles  of  Qeontt 
for  our  present  purposes.     The  principle  may  be  Accepted 
as  a  good  working  principle  until  it  can  be  shown  to  fail,  or 
until  it  can  be  proved  to  be  true  in  all  cases. 

538.  A  little  further  study  of  Fig.  77  would  show  that 
the  altitudes  of  the  .various  similar  triangles  are  related  to 
each  other  just  as  are  the  sides  of  the  corresponding  tri- 
angles ;  thus  the  sides  of  /\  C  F  II  are  f  of  the  OOTrefpond- 
ing  sides  of  /\  C  U  V,  and  the  altitude  of  /\  0  1  II  ifl  also 
{  of  the  altitude  of  i\  C  U  V,  because  the  parallels  D  E, 
F  H,  and  K  R,  etc.,  cut  all  lines  between  0  and  A  B  into 
seven  equal  parts.  Draw  a  figure  lik  ',',.  ami  draw 
the  Altitude  from  C  to  A  B,  and  show  that  it  is  divided  into 
seven  equal  parts  by  the  parallels.  This  fact  gives  additional 
reason  for  believing  the  principle  of  537  to  be  true  ;  for  the 
essential  peculiarity  of  squares,  so  far  as  their  area  is  DOH- 
oernedj  il  that  no  change  can  take  place  in  the  base  of  a 
square,  unless  it  also  takes  place  in  the  side  or  altitude  of 
the  square;  if  the  base  is  made  |  as  large,  the  altitude 
must  also  be  made  |  as  large,  or  the  figure  ceases  to  be  a 
square  :  the  area  of  the  square,  which  is  nothing  but  the 
product  of  the  base  and  altitude,  is  affected  bjf  both  changes, 
and  therefore  becomes  J  of  f  as  large,  or  y  as  large.  Now 
this  same  peculiarity  is  shared  by  similar  triangles  ;  if  the 
base  is  made  |  as  large,  the  altitude  must  also  be  made  f  as 
large,  or  else  the  new  triangle  ceases  to  be  similar  to  the 
old  ;  the  area,  which  is  merely  the  product  of  the  base  and 
the  half-altitude,  is  affected  by  both  changes,  and  becomes 
|  of  |  as  large  or  */  as  large. 

539.  If  similar  triangles  are  like  squares,  they  can  be 
added  like  squares.  Suppose  you  wish  to  add  /\CK  B 
to  A  C  M  X  (Kg.  77)  ;  place  C  K  at  right  angles  to  C  M 
(Fig.  78),  as  if  you  were  to  add  0  K*  to  0  M2 ;  K  If  will  bo 
the  side  of  the  new  triangle  corresponding  to  C  K  and  C  M 
of  the  given  triangles.  Make  a  triangle  with  the  angles  of 
/\  C  K  R,  starting  with  K  M,  Fig.  78,  in  the  position  of 


CONSTRUCTIONAL   GEOMETRY 


107 


C  K,  Fig.  77.  The  triangle  thus  formed  will  be  like  A 
OKR  and  C  M  N  in  shape,  and  will  be  equivalent  to  their 
sum. 

In  Fig.  77  there  is  a  triangle 
that  is  equivalent  to  /\  C  K  R  + 
A  C  M  N.  Find  this  triangle, 
and  compare  it  with  the  triangle 
that  you  have  just  constructed. 

540.  Make  three  pairs  of  simi- 
lar triangles,  and  add  the  trian- 
gles of  each  pair  by  the  method 
described  in  539. 

541.  Can  you  subtract  one  tri- 
angle from  a  larger  triangle  similar 
to  it  ?     Give  two  illustrations. 

542.  Can  you  find  a  triangle  twice  as  large  as  another 
triangle  and  similar  to  it  ?  Suggestion  :  Think  how  you 
double  a  square. 

543.  Make  a  triangle  ABC;  can  you  find  a  similar  tri- 
angle equivalent  to3x/\ABC?   to   4xAABC?to 

5  x  AABC? 

544.  Given  a  triangle,  ABC,  can  you  find  a  similar  trian- 
gle equivalent  to  J  of  A  A  B  C  ?  to  J  of  A  A  B  C  ?  to  f  of 
AABCFtofofAABC? 

545.  Polygons  are  said  to  be  similar  to  each  other  when 
their  corresponding  angles  are  equal,  and  when  their  cor- 
responding sides  have  the  same  relation  to  each  other. 

Make  a  pentagon,  and  then,  with  the  aid  of  your  divid- 
ing tool  and  protractor,  make  another  pentagon  similar  to 
it  and  with  sides  f  as  long  as  those  of  the  first  pentagon. 

546.  A  convenient  method  of  making  one  polygon  simi- 
lar to  another  is  illustrated  in  Fig.  79. 

The  problem  is  to  make  a  pentagon  similar  to  A  B  C  D  E 
with  sides  twice  as  long  respectively. 

Divide  A  B  C  D  E  into  three  triangles  by  the  diagonals 
AC,   AD.     Extend  A  B  to   H,   making  A  H  =  2  A  B. 


Draw  H  K 


B  0  meeting  A  C  extended  in  K. 


In  the  same 


108  ELEMENTARY  AND 

way  draw  K  M  |  C  I)  and  M  N  |  D  E,  completing  the  pen- 
tagon AHKM  N.  The  corresponding  angles  of  tin-  two 
polygons  are  plainly  the  same.  Why  ?  Can  you,  with  the 
help  of  the  principle  of  325,  show  that  II  K,  K  M,  M  N, 


and  N  A  are  respectively  twice  as  long  as  the  corresponding 
.,f  A  BODBf 
How   much  larger  is  each   triangle  of  A  II  K  M  N  than 
the    corresponding   triangle   of   A  B  C  D  K  !'     How    mu.h 
larger  is  A  II  K  M  N  than  ABOD  E? 

547.  Make  a  hexagon,  and  then  make  a  similar  hexagon 
with  sides  \  as  long  as  those  of  the  first,  using  the  method 
described  in 

Compare  the  areas  of  the  corresponding  triangles  and  of 
the  two  hexagons. 

548.  Do  similar  polygons  resemble  squares  in  respect  to 
their  surfaces? 

549.  Oan  yon  make  a  parallelogram  twice  as  large  as  a 
given  parallelogram,  and  similar  to  it? 

550.  (an  you  make  a  pentagon  similar  to  those  in  Fig. 
.ml  equivalent  to  their  sum  P 

551.  (an  you  make  a  pentagon  similar  to  those  in  Fig. 
79,  and  equivalent  to  their  difference  ? 


CONSTRUCTIONAL   GEOMETRY 


109 


552.  Can  you  make  a  hexagon  $  of  a  given  hexagon  ?  f 
of  a  given  hexagon  ?  \  of  a  given  hexagon  ?  In  each 
case  the  hexagon  must  be  similar  to  the  given  hexagon. 

553.  Can  you  make  a  circle  equivalent  to  the  sum  of  two 
given  circles  ? 

554.  Can  you  subtract  one  circle  from  another,  giving  the 
difference  in  the  form  of  a  circle  ? 

555.  Can  you  change  a  circular  ring,  or  band,  into  a 
circle  ? 

556.  Can  you  make  a  circle  twice  as  large  as  another  cir- 
cle ?  three  times  as  large  ?  four  times  as  large  ? 

557.  One  circular  hole  was  found  to  have  §  of  the  diam- 
eter of  another ;  how  much  smaller  than  the  second  hole 
was  the  first  hole  ? 

558.  If  you  wish  to  double  the  amount  of  light  coming 
through  a  circular  window,  should  you  double  the  diameter 
of  the  window  ? 

559.  Draw  a  circumference,  Fig.  80,  and  with  your  di- 
viding-tool divide  the  diameter  into  five  equal  parts ;  draw 
semicircumferences  on  the  di- 
ameters AB,  AG,  AD,  and 
A  E  above  the  diameter  A  F, 
and  semicircumferences  on  the 
diameters  B  F,  C  F,  D  F,  and 
E  F  below  the  diameter  A  F. 

What  part  of  the  semicircle 
A  H  F  is  A  K  B  ? 

What  part  of  the  whole  circle 
is  semicircle  A  K  B  ? 

What  part  of  the  whole  circle 
is  semicircle  B  M  F  ? 

The     horn  -  shaped     figure 
A  K  B  F  T  A  is  made   up    of 
the  semicircles  A  K  B  +  A  T  F 
the  whole  circle  is  the  horn  ? 

If  the  corresponding  lines  of  similar  figures  have  the 
same  relation  to  each  other,  what  part  of  the  semicircum- 


B  M  F.     What  part  of 


110  ELEMENTARY  AND 

ference  A  II  F  should  the  semicircumference  A  K  B 
What  part   of   the  semicircumference  A  II  F  should    the 
semicircumference  B  M  F  be  ? 

Which  is  the  longer  path  from  A  to  F,  the  path  by  the 
semicircumference  A  T  F,  or  that  by  the  two  semicircum- 
ferences  A  K  B  and  B  M  F  ? 

560.  By  adding  and  subtracting  the  right  semicircles  in 
Fig.  80,  the  surface  of  the  strip  A  L  C  N  F  M  B  K  A  <  an 
be  found.  Can  you  select  the  right  semicircles,  and  can 
you  estimate  what  part  of  the  whole  circle  each  semicircle 
is,  and  what  part  of  the  whole  circle  the  strip  is  ?  Can  you 
estimate  the  perimeter  of  the  strip  ? 

561.  In  559  you  saw  thai  the  horn  AKBFTA  is 
equivalent  to  the  semicircles  A  K  B  +  A  T  P  —  B  M  P  :  can 
you  find  one  semicircle  that  shall  be  equivalent  to  A  K  B 
+ATF-BMF?    Bee  568  and  654. 

Make  a  circle  ami  divide  the  diameter  into  eight 
parti ;  draw  semicircumferences,  as  in  Fig.  80,  cutting  the 
circle  into  eight  strips.  Try  to  show  that  the  circle  is  thus 
divided  into  eight  equivalent  strips. 

563.  Can  you  find  a  circle  one-half  of  a  given  circle  ? 
one-third  of  a  given  circle  ?  two-fifths  of  a  given  ein  1.   \ 

."••'.I.  Review  carefully  Bectioni  XV.,  XVI..  and  XVII. 
(in  several  lessons  if  necessary). 

565.  Write  a  sketch  of  the  square,  describing  how  squares 
can  be  added  and  subtracted  ;  how  they  can  be  enlarged  any 
number  of  times  without  losing  their  shape;  and  how  they 
can  be  divided  into  as  many  small  squares  as  may  be  de- 
sired. Tell,  also,  what  other  figures  can  be  added,  sub- 
tracted, enlarged,  or  diminished  by  the  methods  used  for 
squares. 


CONSTRUCTIONAL   GEOMETRY  111 


SECTION  XVIII.     CIRCLES   AND  INSCRIBED 
ANGLES. 

566.  A  circle  is  a  figure  bounded  by  a  line,  called  the 
circumference,  every  point  of  which  is  equally  distant 
from  a  point  within,  called  the  centre.  The  sign  for  circle 
is  0,  plural  ©  ;  and  that  for  circumference  is  O,  plural 
©.  You  have  already  learned  that  you  can  measure  an 
angle  by  placing  its  vertex  at  the  centre  of  a  circle  whose 
circumference  has  been  divided  into  three  hundred  and 
sixty  equal  parts,  and  by  noting  the  number  of  these  parts, 
or  degrees,  between  the  sides  of  the  angle.  Any  portion  of 
a  circumference  is  called  an  arc,  and  the  angle  formed 
by  the  radii  through  the  ends  of  an  arc  is  said  to  sub- 
tend the  arc.     Thus  in  Fig.  81 

/BOA  subtends  the  arc  B  A, 
written  BA.  The  number  of  de- 
grees in  the  arc  B  A  is  the  same 
as  the  number  of  degrees  in  the 
angle  B  0  A  at  the  centre  ;  if 
BA  =  60°,  /  B  0  A  =  60°  ;  if 
BA  =  50°,  I  B  0  A  =  50°  ;  if 
BA  =  x°,  /_  B  0  A  =  x° ;  so 
that  an  angle  at  the  centre  of  a 
circle  has  the  same  name  as  the 
arc  which  it  subtends.  This  prin- 
ciple is  simply  a  restatement  of  the  principle  of  the  pro- 
tractor. 

567.  In  using  a  protractor  you  have  always  placed  the  cen- 
tre of  the  instrument  upon  the  vertex  of  the  angle.  The  pe- 
culiarities of  circles  enable  us  often  to  measure  angles  with- 
out a  protractor,  and  also  to  measure  angles  whose  vertices 
are  not  at  the  centre  of  the  circle.  In  Fig.  82  A  0  E  is  a 
diameter,  and  AB  =  60°.  Can  you,  without  a  protractor, 
give  the  value  in  degrees  of  all  of  the  angles  in  the  figure  ? 


112 


ELEMENTARY  AND 


Can  you  see  any  simple  relation  between   /   BE  A  ami 

BA  P 

5G8.  Imagine,  or  draw,  an  arc  A  C  =  70°,  and  the  diam- 
eter A  E.  Can  you  tell  how  many 
degrees  there  are  in  the  angle 
C  E  A  ?  Try  arcs  of  <K)°,  75°, 
30°,  120°,  and  150°,  from  A; 
and,  after  joiuing  the  end  of 
each  arc  with  B,  find  the  value 
of  each  angle  formed  at  K  with 
the  line  E  A.  Draw  a  figure 
like  Fig.  82  for  each  case,  mak- 
ing the  arc  A  B  the  proper 
length  each  time. 

An  angle,  placed  as  an- 
gle   B   B  A    is   place*  1    in    Fig. 

B8,  1  ith  its  vertex  on  the  ei re u inference,  and  with  its  sides 
running  to  other  points  of  the  circumference,  forming 
chords,  is  called  an  inscribed  angle.  The  Experiment!  of 
and  5G8  suggest  the  principle  that  inscribed  angles, 
which  have  a  diameter  for  one  tide,  subtend  arcs  of  twice  their 
own  number  of  degrees.  Try  to 
prove  this  principle  by  showing 
that  if  /  E  =  *°,  All  must  be 
equal  to  &b°  Fig.  83.  Draw  II 
O  ami  oompare  j_  II  0  A  with 
/  x.  How  is  /_  H  0  A  related 
to   II  \  I 

570.  What  is  the  largest  in- 
scribed angle  that  you  can  draw 
with  one  side  a  diameter? 

Prom  one  end  of  a  diameter 
draw  chords  that  shall  make 
angles  of  60°,  45°,  30°,  22J°, 

and  75°  with  the  diameter,  using  your  compasses  and  ruler 
only. 


Warn,  ^ 


CONSTRUCTIONAL  GEOMETRY 


113 


Fig.  84. 


571.  Place  the  inscribed  angle  x,  so  that  the  centre  of 
the  circle  shall  come  between  the 

two  chords  which  form  its  sides, 
as  in  Fig.  84.  Will  AH  still  be 
equal  to  2x°  ?  Draw  the  diame- 
ter EOK,  dividing  /  x  into 
two  pieces,  each  of  which  has  a 
diameter  for  one  side.  If  you 
give  to  piece  A  E  K  the  name  y, 
the  piece  KEH  will  be  x  —  y. 
AK  =  2y° ;  KH  =  2x°  -  2y°  ; 
.  • .  HA  =  2^°  +  2x°  -  2y°  = 
2x°,  so  that  HA  =  2x°  as  before. 

If  /  x  be  inscribed,  as  in  Fig.  85,  in  such  a  way  that 
both  chords  come  on  the  same 
side  of  the  centre,  will  the  arc 
A  H  still  contain  twice  as  many 
degrees  as  the  angle  x  ?  Draw 
EOK,  the  diameter,  and  call 
/   A  E  K,    /    y,    as    before ; 

KA  =  2if,  and  HK  =  2x°  + 
2y°.  .  • .  HA  =  HK  -  KA  = 
2x°  +  2ij°  —  2y°  =  2x°  as  be- 
fore. We  therefore  can  say  that 
any  inscribed  angle  subtends  an 
arc  of  twice  its  own  number  of  degrees. 

572.  How  large  an  arc  must  an  inscribed  angle  of  90° 
subtend  ?  an  angle  of  22J°  ?  an  angle  of  150°  ? 

573.  Several  arcs  of  a  circumference,  when  measured, 
were  found  to  be  respectively,  70°,  83°,  90°,  340°,  and 
180°  ;  how  large  was  each  inscribed  angle  that  subtended 
these  arcs  ? 

574.  An  inscribed  angle  is  said  to  stand  on  the  arc  which 
it  subtends.  Can  you  say  why  all  inscribed  angles  in  one 
circle  which  stand  on  the  same  arc  must  be  equal  ?     Make 

8 


Fig.  85. 


114 


ELEMENTARY  AND 


a  circle,  and  inscribe  any  angle  A  E  II.  Make  six  other 
angles  in  the  same  circle  equal  to  /  A  E  H,  using  your 
ruler  only. 

575.  Make  a  circle,  and  make  six  right  angles  by  means 
of  this  circle,  using  no  tool  except  your  ruler  to  make  the 
angles.  Compare  the  results  here  with  what  you  have 
already  learned  about  the  middle  point  of  the  hypotenuse 
of  a  right  triangle. 

576.  Can  you  inscribe  a  regular  hexagon  in  a  circle  ? 
Can  you  determine  the  number  of  degrees  in  each  angle  of 
the  hexagon  by  the  arc  subtends  1  P 

577.  Can  you  inscribe  a  regular  octagon  and  find  the 
number  of  degrees  in  each  angle  ? 

578.  Inscribe  an  angle  of  75°  (/  A  E  II,  Fig.  86). 
Choose  any  point  V  in  the  arc  A  II  and  draw  the  chords 
II  V  and  A  V  completing  the 
quadrilateral  A  V  II  E.  Can 
you  tell  the  number  of  degrees 
in  the  angle  at  V  P  How  is 
/  II  V  A  related  to  /  E  ?  Can 
you  give  the  value  of  the  angles 
at  II  an.l  A  ?  Can  you  tell 
the  number  of  degrees  in  their 
sum  ? 

579.  Can     yon     show     what 
must  always  Ik*  the  relation  be- 
tween   the   opposite    angles  of 
an  inscribed  qaadri late ral  ?    Is   there   any    parallelogram 
whose  opposite  angles  have  this  required  relation  P 

580.  Can  you  make  a  cinu inference  pass  through  the 
four  comers  of  a  rhomboid  ?  of  a  rhombus  ?  of  a  rectangle  ? 
of  a  square  ? 

581.  Draw  a  circle,  Fig.  87,  and  then  draw  two  paral- 
lel chords,  A  E  and  II  K.  By  the  aid  of  A  K  can  you 
show  that  the  are  A  II  is  equal  to  the  arc  K  E,  and  thus 
establish  the  principle  that  parallel  chords  intercept  equal 
arcs? 


CONSTRUCTIONAL  GEOMETRY 


115 


Fig.  87. 


582.  Any  chord  divides  a  circle  into  two  segments ;  for 
example,  A  E,  Fig.  87,  divides  the  circle  into  the  two  seg- 
ments AHKE,  and  AVE. 
Segments  are  named  by  the  num- 
ber of  degrees  in  their  arcs ;  thus 
a  segment  of  200°  is  one  whose 
arc  is  200°  ;  a  semicircle  is  a  seg- 
ment of  180°.  If  one  of  the 
segments  into  which  a  chord  di- 
vides a  circle  is  120°,  what  is 
the  other  segment  ?  An  angle 
is  inscribed  in  a  segment  when 
its  vertex  is  in  the  arc  of  the 
segment,  and  when  the  sides  of 
the  angle  pass  through  the  ends  of  the  chord  that  makes 
the  segment.  Thus  the  angle  A  F  E  would  be  an  angle 
inscribed  in  the  segment  AVE,  Fig.  87. 

583.  An  angle  A  T  E  is  inscribed  in  a  segment  of  100°  ; 
what  is  the  value  of  the  arc  A  E  on  which  the  angle  stands  ? 
of  the  arc  A  T  E  ?  of  the  angle  ATE? 

584.  How  large  a  segment  is  required  that  an  angle  of 
120°  may  be  inscribed  in  it  ? 

585.  What  can  you  say  of  all 
angles  inscribed  in  the  same  seg- 
ment ?     Why  ? 

586.  If  an  angle  is  inscribed  in 
a  semicircle,  what  must  it  be  ? 

587.  Of  two  inscribed  angles, 
which  is  inscribed  in  the  larger 
segment,  the  larger  angle  or  the 
smaller  one  ? 

588.  If  a  line  merely  touches 
a  circle  at  one  point,  so  that, 
when  extended  in  either  direc- 
tion, it  does  not  pass  within,  or  out,  the  circle,  it  is  called 
a  tangent  to  the  circle  ;  the  point  at  which  it  touches  the 
circle  is  called  the  point  of  contact,     In  Fig.  88,  T  A  is  a 


116  ELEMENTARY  AND 

tangent  to  the  circle  with  its  point  of  contact  at  T.  Can 
you  explain  why  the  name  tangent  was  selected  for  this 
line  ? 

589.  Draw  any  radius,  as  0  T,  Fig.  88,  and,  at  its  ex- 
tremity T,  draw  a  line  T  A  X  0  T.  Do  this  without  ex- 
tending 0  T,  if  you  can.  Can  you  show  that  T  A  can- 
not touch  the  circle  in  any  point  except  T,  so  that  it  must 
be  a  tangent  ?  Suggestion :  Let  A  stand  for  any  point 
of  the  line  except  T,  and  show  what  kind  of  triangle 
OTA  would  necessarily  be,  if  A  were  on  the  circumference, 
and  also  what  would  be  the  value  of  the  /  O  A  T  in  such 
a  case.  Is  such  a  triangle  as  ^  0  T  A  would  become,  if 
A  were  on  the  circumference,  a  possible  triangle  P     Why  P 

590.  Choose  several  points  on  the  circumference  of  a  cir- 
cle, and  draw  tangents  touching  the  circle  at  these  points 
respectively. 

591.  Thi  angle  A  T  II,  Fig.  89,  made  by  a  tangent  T  H 
and  a  chord  T  A  through  the  point  of  contact  is  not  called 

an  inscribed  angle,  but  like  an  in-  

scribed    angle    it   contains  only        yr  N. 

one  half  as  many  degrees  as  the      /  \ 

arc  T  E  A  subtended  by  the  chord    /  \ 

T  A.      Study  ,\  T  V  0  formed  /  -— A| 

by  drawing  the  radius  OT  and   I  Q^  7\ 

0  F  the   bisector  of  the  angle   \  [    \      /  J 
T  O  A.     What  kind  of  triangle    \  rVs     / 
is  A  T  O  A  P     Select   as   many      \             !     /  y<£ 
pairs  of  equal  angles  as  yon  can          ^^^  v^^ 
from  your  knowledge  of  the  nat-                      T  FT 
nre  of  ^  T  0  A  and  from  the                   *»•  »• 
method  of  drawing  0  F.     How  many  degrees  in  /  FTO  + 

1  T  0  F  ?  How  many  degrees  in/FTO+/FTH? 
Why  must  /  V  THor  /  A  T  n  =  /  T  0  F  ?  Now 
/  TOF  has  as  many  degrees  as  'ffe,  or  one  half  as  many 
degrees  as  TA.  Hence  /  AT  II  has  one  half  as  many  de- 
-iv.-  ;i<  '1'  \.  or  an  ant^le  formed  by  a  tangent  and  a  chord 


CONSTRUCTIONAL  GEOMETRY  117 

through  the  point  of  contact  contains  one  half  as  many  de- 
grees as  the  intercepted  arc. 

592.  How  is  any  angle  inscribed  in  the  larger  segment 
formed  by  T  A,  Fig.  89,  related  to  /ATH?  How  is  any 
angle  inscribed  in  the  smaller  segment  A  E  T  related  to 
/  A  T  H  ? 

593.  A  chord  A  B  intercepts  an  arc  of  80°.  How  many 
degrees  are  there  in  the  angle  between  A  B  and  the  tan- 
gent at  A  ?  How  many  degrees  in  the  angle  between  B  A 
and  the  tangent  at  B  ? 

594.  Draw  the  tangent  to  the  circle  of  Fig.  89  at  A  as 
well  as  at  T.  Can  you  give  any  reason  for  thinking  that 
the  angle  between  the  tangent  at  T  and  T  A  is  the  same  as 
the  angle  between  the  tangent  at  A  and  AT? 

595.  What  kind  of  triangle  will  be  formed  by  a  chord 
and  the  two  tangents  at  its  extremities  ? 

596.  Draw  a  circle  with  centre  0,  Fig.  90,  and  choose  a 
point,  P,  outside  the  circle.  It  is  required  to  draw  a  line 
from  P  that  shall  be  tangent  to  the  circle. 


If  0  A  P  is  a  right  angle,  A  P  will  be  tangent  (589). 
Study  Fig.  90,  and  try  to  explain  what  principles  were 
used  to  determine  the  points  A  and  B  that  /sOAP  and 


118 


ELEMENTARY  AND 


OBP  might  be  90°.     Can  you  see  what   help  line  was 
drawn  first  ? 

597.  Make  four  different  circles,  and  draw  tangents  to 
them  from  points  selected  outside  of  the  circles. 

598.  If,  in  Fig.  90,  0  A  is  5  in.,  and  0  P  is  13  in.,  how 
long  is  A  P  ?  What  principle  enables  you  to  answer  this 
question  ? 

A  ball,  whose  radius  is  7  in.,  rests  in  a  hollow  cone 
so  that  its  points  of  contact  with  the  cone  are  24  in. 
from  the  apex  of  the  cone  ;  how  far  is  the  centre  of  the  ball 
from  the  apex  ? 

600.  If  the  radius  of  the  earth  is  4000  miles,  how  far 
out  to  sea  does  the  horizon  of  a  person  extend,  if  he  is  C  ft. 
tall,  and  if  he  is  standing  at  the  water's  edge  ? 

801.  How  far  can  the  light  of  a  light-house  105  ft.  high, 
be  seen  from  the  surface  of  the  water  ? 

H«>w  far  can  the  same  light  be  seen  from  a  point  in  the 
rigging,  thirty  feet  above  the  surface  of  the  water  ?  Con- 
sider the  radius  of  the  earth  4<>oo  miles  in  both  cases. 

I  >aw  a  circle  with  centre  at  0 ;  then  take  P  at  a 
distance  from  <>  equal  to  the  diameter  of  the  circle.  Draw 
the  tangents  PA  and 
P  B,  and  the  chord 

of  contact.  A  I'.. 
Can  you  give  the 
number  of  degrees 
in  all  the  angles 
formed,  with  rea- 
sons in  each  case  ? 
How  many  degrees 
in  the  arc  A  Q  B  ? 
Why  ?  In  the  arc 
AOH?  Why  ?  If 
the  radius  of  the  first  circle  is  8  in.,  can  you  find  the  lengths 
of  all  the  lines  in  the  figure,  curved  and  straight  t 

603.  What  part  of  the  circumference  is  \Q\\.  Fig,  '.'1  P 
What  part  of  the  earth's  surface  would  a  mau  see  were  he  to 


A^-- 


Flo.  91. 


CONSTRUCTIONAL   GEOMETRY  119 

ascend  4000  miles  above   the   earth's  surface  ?     How   far 
would  he  have  to  go  to  see  one  half  of  the  earth's  surface  ? 

604.  In  Fig.  91,  if  you  call  angle  A  P  B,  x°,  can  you  say 
how  many  degrees  there  must  be  in  the  arc  AQB  ? 

Two  tangents  make  an  angle  of  30°  with  each  other; 
how  long  an  arc  do  they  inclose  ? 

If  you  want  tangents  to  meet  at  an  angle  of  45°,  how 
many  degrees  apart  must  you  take  the  points  of  contact  ? 

605.  Can  you  state  a  principle  concerning  the  relation  be- 
tween the  angle  which  two  tangents  form  and  the  arc  in- 
cluded between  the  points  of  contact  ? 

606.  In  Fig.  91,  assuming  that  the  radius  is  8  in.,  can 
you  find  the  number  of  square  inches  inclosed  between  the 
tangents  and  the  arc  AQB?  Find  also  the  area  of  the 
oval  A  O  B  Q  ;  of  the  smaller  segment  cut  off  by  A  B  ;  and 
of  the  triangle  A  O  B. 


607.  Three  circles  of  equal  size  with  a  radius  of  6  in.  are 
placed  in  contact  as  in  Fig.  92. 

Can  you  find  the  area  of  the  space  between  the  circles  ? 


120 


ELEMENTARY  AND 


What  is  the  area  of  the  parts  of  the  circles  without  the  tri- 
angle of  centres  0  P  Q  ? 

608.  Draw  a  circle,  and  draw  tangents  to  this  circle  at 
points  D,  F,  and  L,  Fig.  93,  which  are  120°  apart,  thug 
forming  /\ABC.  Draw 
ADFL,  and  also  A  H  P  E, 
whose  vertices  are  at  the  mid- 
dle points  of  the  arcs  F  L,  L  D, 
and  1)  F,  respectively.  Can 
you  account  for  the  shape  of 
the  triangles  formed  P  How 
much  smaller  than  the  outside 
triangle  is  the  smallest  tri- 
angle ?  (Recall  the  principle 
about  the  surfaces  of  similar 
triangles.)  What  part  of  the 
intermediate  triangle  is  the 
hexagon  ?     Find  the  number 

of  square   inches   in  each    triangle  and   in  the  hexagon,  if 
A   B  =  (»  in.      Find  also  the  area  <»f  the  circle. 

<'.<>!*.  Can  yon  make  a  circum- 
ference pan  through  two  points, 
A  ami  B  P  How  many  cireuni- 
terencet  P  What  can  you  say  of 
the  position  of  the  centres  of  all 
the  circumferences  that  pass 
through  A  and  B  ? 

610.  Can  you  make  a  circum- 
ference pass  through  any  three 
points,  A,  B,  and  II  P  Bow 
many  circumferences  can  you 
draw  through  A,  B,  and  II  P 
Can  you  put  A,  B,  and  11  in 
such  positions  that  a  circumference  cannot  be  drawn 
through  them  ? 

Gil.   Under  what  circumstances  can  you  draw  a  circum- 
ference through  four  points  ?    Can  you  recall  any  quadri- 


Fiu.  M. 


CONSTRUCTIONAL   GEOMETRY 


121 


lateral  through  whose  corners  you  could  draw  a  circumfer- 
ence ? 

612.  Make  a  line,  and  a  point  without  the  line  ;  can  you 
draw  a  circumference  that  shall  merely  touch  the  line,  and 
at  the  same  time  pass  through  the  point  ?  How  many  cir- 
cumferences fulfilling  these  conditions  can  you  draw  ? 

613.  Can  you  draw  a  circumference  that  shall  be  tangent 
to  the  line  of  612  at  a  chosen  point,  and  at  the  same  time 
pass  through  the  point  without  the  line  ? 

614.  Can  you  make  a  circumference  with  a  given  radius, 
touch  a  given  line,  and  pass  through  a  given  point  ?  Can 
the  radius  be  made  too  long  ?     Can  it  be  made  too  short  ? 

615.  Can  you  make  a  circumference  pass  through  the 
vertices  of  a  scalene  triangle  ? 

616.  Using  some  round  object  as  a  guide,  draw  a  circle  ; 
can  you  find  the  centre  of  the  circle  ? 

617.  Draw  free  hand  a  curved  line  ;  can  you  tell  whether 
or  not  the  curve  is  a  piece  of  a  circumference  ? 


Fig.  95. 


618.  Can  you  make  a  circle  touch  two  intersecting  lines  ? 
How  many  answers  can  you  obtain  for  one  pair  of  inter- 
secting lines  ? 

619.  Can  you  make  a  circle  touch  the  three  sides  of  a 
triangle  ? 

620.  Can  you  make  a  circle  touch  two  lines  that  do  not 


122 


ELEMENTARY  AND 


meet  within  the  boundaries  of  the  paper,  although  they  are 
not  parallel  lines  ? 

621.  At  two  points,  A  and  B,  of  a  line  (Fig.  95)  draw 
two  circles  tangent  to  the  line,  using  different  radii  of 
such  lengths  that  the  circles  will  have  no  point  in  common. 

Draw  the  radii,  A  P  and  B  Q,  and  join  the  centres  of  the 
circles  by  the  line  P  Q. 

What  kind  of  quadrilateral  is  A  B  Q  P  ? 

llow  many  of  its  angles  do  you  know  ? 


Yu.    OC, 


Draw  a  line  Q  II  parallel  to  B  A.  Describe  the  two 
figures  into  which  Q  II  divides  the  quadrilateral  A  B  Q  P. 
If  the  radial  of  the  larger  circle,  Fig.  95,  is  8  in., 
and  that  of  the  smaller  is  3  in.,  and  if  the  centres  of  the 
circles  are  I'.i  in.  apart,  can  you  find  the  length  of  the  com- 
mon tangent  AB?  of  HP?  of  HQ? 

G23.  Can  you  make  a  trapezoid,  knowing  the  two  bases 
and  one  leg,  so  that  the  bases  will  make  right  angles  with 
the  unknown  leg  ?  Suggestion  :  Try  to  reproduce  Fig.  95 
from  three  lines  equal  in  length  to  P  Q,  A  P,  and  B  Q. 


CONSTRUQTIONAL   GEOMETRY  123 

624.  Can  you  draw  a  line  that  shall  touch  two  given 
circles"? 

Study  Fig.  96,  where  the  two  given  circles  have  their 
centres  at  0  and  P,  and  where,  by  the  aid  of  the  circle 
with  0  P  as  diameter,  the  tangent  T  E  was  determined. 
Compare  the  quadrilateral  OTEP  with  the  quadrilateral 
P  A  B  Q  in  Fig.  95. 

In  Fig.  96  a  line,  R  H,  is  found,  which  touches  the  two 
circles  and  passes  between  them.  Can  you  see  how  points 
F  and  R  were  determined  when  this  interior  tangent  was 
drawn  ? 

625.  Make  two  circles  and  draw  an  interior  tangent  to 
them.     Repeat  the  problem  with  different  circles  twice. 


SECTION  XIX.     VOLUMES. 

626.  In  367  and  the  following  exercises  you  learned  that 
the  product  of  two  lines  represents  a  surface ;  that  the 
simplest  surface  that  you  can  form,  in  multiplying  two 
lines  together,  is  the  rectangle ;  but  that  this  surface  can 
be  moulded  into  many  other  forms. 

627.  Multiply  a  line  7  cm.  long  by  one  4  cm.  long ;  the 
rectangle,  representing  their  product,  contains  28  sq.  cm. 
See  rectangle  A  B  C  D  (Fig.  97),  where  the  28  sq.  cm.  are 
outlined.  Draw  a  line,  B  F,  5  cm.  long,  that  shall  be  per- 
pendicular to  both  B  A  and  B  C,  and  complete  the  rectan- 
gular parallelopiped  A  B  C  D  E  F  K.  If  B  S  is  taken 
equal  to  one  centimeter,  and  if  the  figure  is  divided  as 
represented,  what  are  the  dimensions  of  the  solid  S  B  M. 
What  is  its  name  ?  How  many  solids  like  it  could  be  made 
out  of  the  solid  H  S  T  D,  which  forms  the  top  layer? 
How  many  layers  like  HSTD  could  be  cut  from  the 
whole  solid?  How  many  cubic  centimeters  (ccm.)  could 
be  cut  from  the  whole  solid  ?  How  do  you  obtain  the 
answer  to  the  last  question  ? 


124 


ELEMENTARY  AND 


628.   How  many  ccm.  of  wood  are  there  in  a  rectangu- 
lar piece  8  cm.  long,  5  cm.  wide,  and  3  cm.  thick. 

How  many  cubic  inches  (cu.  in.)  of  ice  could  be  cut  from 


2l 


.^z^^S2^ 


H 


f^Ba 


Flo    9T. 


a  rectangular  block  of  ice  24  in.  long,  15  in.  broad,  and  8 
in.  thick  ? 

620.  Just  as  you  saw  that  the  product  of  two  lines  rep- 
resents a  surface,  you  can  now  see  that  the  product  of  three 
lines  represents  a  solid.  The  simplest  solid  that  you  can 
form,  in  multiplying  three  lines  together,  is  the  rectangu- 
lar oblong,  or  rectangular  parallelopiped  formed  by  using 
the  three  lines  for  the  edges  which  are  at  right  angles  to 
each  other. 

The  product  of  two  of  the  lines  gives  the  base  of  the  ob- 
long ;  the  third  line  is  the  altitude  of  the  oblong.  It  is  not 
necessary  to  form  a  mental  picture  of  the  small  cubes  into 
wliieh  the  oblong  may  be  divided  eaeli  time  that  you  mul- 
tiply three  lines  ;  in  fart  it  is  better  to  think  of  the  space 
occupied  by  the  solid  as  a  whole. 

630.  Make  four  sets  of  three  lines  each,  and  draw  the 
rectangular  oblong  represented  by  the  product  of  the  three 
lines  of  each  set. 


CONSTRUCTIONAL  GEOMETRY 


125 


631.  Draw  the  oblong  of  Fig.  97  again,  this  time  with- 
out the  additional  dividing  lines.  You  recall  that  its  dimen- 
sions are  7  cm.  by  4  cm. by  5  cm.  Imagine  a  piece  cut  from 
the  solid  at  the  left  end  along  the  line  VTM,  and  imagine 
this  piece  transferred  to  the  right  end,  with  the  face  DAE 


Fig.  98. 


placed  upon  its  equal,  C  B  F  K.  Will  the  new  solid 
V  T  M  E  L  P  H  occupy  more  or  less  space  than  the  old  ob- 
long ?  Will  its  dimensions  be  changed  ?  (It  is  supposed 
that  the  cutting  is  done  without  waste.)  What  shape  will 
the  faces  of  the  new  solid  have  ?  Will  the  twelve  edges  of 
one  solid  be  any  longer  than  the  twelve  edges  of  the  other  ? 
Why  ?  Which  of  the  new  faces,  if  any,  cover  any  larger 
surface  than  the  corresponding  faces  of  the  oblong  ? 
Which  solid  has  the  larger  upper  base  ?  Which  lines  must 
you  multiply  together  to  obtain  the  upper  base  VTHP? 
What  lines  must  you  multiply  together  to  obtain  the  solid 
VTMRLPH? 

632.  The  solid  VTMRLPH  (Fig.  98),  is  called  a 
paralellopiped.  Do  you  understand  why  ?  If  possible  se- 
cure two  blocks  of  wood,  one  a  rectangular  parallel  opiped, 
and  the  other  a  parallelopiped,  both  of  which  have  been  cut 
from  one  long  rectangular  block.     The  two  parallelopipeds 


126  BLBMRNTARY  AND 

should  have  one  edge  (corresponding  to  edge  D  C,  Fig.  98) 
the  same  in  both.  After  coating  the  blocks  with  parajfine, 
that  they  may  not  absorb  water,  sink  first  one  and  then  the 
other  in  a  vessel  of  water,  noting  carefully  how  high  the  water 
rises  in  the  vessel  each  time.  Does  the  experiment  confirm 
your  answers  to  the  questions  of  631  which  relate  to  the  vol- 
ume of  the  solids  ?  Try  the  experiment  of  weighing  the  two 
blocks,  and  see  whether  the  result  agrees  with  that  of  your 
first  experiment. 

633.  Measure  in  centimeters  the  edges  of  the  rectangu- 
lar block  used  in  632,  estimate  the  volume  of  the  block  in 
cubic  centimeters  as  in  628,  and  then  verify  your  estimate 
by  pouring  from  a  vessel  graded  in  cubic  centimeters 
enough  water  to  raise  the  level  of  the  water  in  the  vessel 
used  in  632  to  the  same  height  to  which  it  was  raised  by 
sinking  the  block. 

634.  In  cutting  the  parallelepiped  from  the  rectangular 
block,  Fig.  98,  do  you  alter  the  size  of  the  parallelopiped  by 
changing  the  directions  in  which  you  draw  the  lines  V  X 
and  T  M  to  guide  your  rutting  ?  Do  you  change  the  area 
of  the  upper  face  as  you  change  the  direction  of  V  T  ? 
Can  you  draw  V  T  and  T  M  in  such  a  way  that  the  paral- 
lelopiped  will  have  four  of  its  faces  rectangles  ?  two  of  its 
faces  rectangl- 

.  Can  you  multiply  three  lines  together  so  that  the 
product  will  be  a  rectangular  parallelepiped?  i  parallele- 
piped with  four  faces  rectangles  ?  with  two  faces  rectangles  ? 
with  no  faces  rectangles  ? 

636.  Can  you  give  any  conditions  under  which  two  paral- 
lelopipeds  of  different  shapes  will  occupy  the  same  amount 
of  space?  Think  of  the  corresponding  conditions  for 
equivalent  parallelograms. 

I  haw  any  parallelopiped  that  is  not  rectangular,  and 
draw  the  lines  whose  product  represents  the  volume  of  the 
solid.  Form  the  rectangular  parallelopiped  that  is  equiva- 
lent to  the  first  solid. 

G38.  Take  a  block  of  wood  in  the  form  of  an  oblique- 


CONSTRUCTIONAL   GEOMETRY  127 

angled  parallelopiped  ;  find  the  lines  whose  product  repre- 
sents the  volume  of  the  parallelopiped,  measure  these  lines 
in  centimeters,  estimate  the  volume  of  the  solid,  and  test 
your  answer  by  the  method  of  633. 

639.  The  principle  to  be  derived  from  Articles  627-638 
is  that  a  parallelopiped  is  the  product  of  its  three  di- 
mensions, or  from  another  point  of  view,  the  volume  of  a 
parallelopiped  is  the  product  of  its  base  by  its  altitude. 
Just  as  a  rectangle  can  be  moulded  into  any  parallelogram 
which  has  the  same  base  and  altitude  as  the  rectangle,  a 
rectangular  parallelopiped  can  be  moulded  into  any  oblique 
parallelopiped  which  has  the  same  base  and  altitude  as  the 
rectangular  parallelopiped. 

640.  Cut  a  rectangular  block  of  wood  into  two  pieces  by 
sawing  along  two  diagonally  opposite  edges.  The  result- 
ing pieces  are  called  triangular  prisms ;  the  two  triangular 
faces  form  the  bases  of  the  prism,  and  the  three  faces  in 
the  shape  of  a  parallelogram  form  the  lateral  surface. 
When  a  rectangular  parallelopiped  is  cut  into  two  triangu- 
lar prisms,  as  in  this  case,  the  prisms  are  called  right 
prisms  ;  for  a  right  prism  is  one  whose  lateral  edges  are  at 
right  angles  to  the  two  bases.  Do  the  two  right  prisms 
formed  in  the  experiment  occupy  the  same  amount  of 
space  ?  Test  your  answer  to  the  question  by  weighing  the 
two  prisms,  or  by  sinking  them  in  a  vessel  of  water,  and  by 
noting  the  height  to  which  the  water  rises  in  each  case. 

641.  Can  you  find  the  three  lines  whose  product  repre- 
sents the  volume  of  one  of  the  prisms  of  640  ?  Can  you 
draw  the  rectangular  parallelopiped  that  will  occupy  the 
same  space  as  the  prism  ? 

64^.  Can  you  show  that  the  product  of  two  of  the  three 
lines  found  in  641  represents  the  base  of  the  prism,  and 
that  the  third  line  is  the  lateral  edge,  and  at  the  same  time 
the  altitude  of  the  prism  ? 

In  doing  this  you  illustrate  an  important  principle  of 
Geometry  :  The  volume  of  a  right  triangular  prism  is  the 
product  of  its  base  by  its  altitude. 


128 


ELEMENTARY  AND 


643.  Imagine  the  oblique  parallelopiped  ( A  B  C  D  E  I   1 1 . 
Fig.  99),  none  of  the  faces  of  which  are  rectangles,  to  be  cut 


Flo.  99. 

into  oblique  triangular  prisms  by  a  plane  through  the  di- 
agonally opposite  edges  A  B  and  £  D.  Imagine  also,  that, 
before  the  pieces  are  separated,  the  parallelopiped  is  cut 
along  the  lines  M  K  and  K  L,  which  have  been  drawn  at 
right  angles  to  E  II.  Transfer  the  portion  cut  off  at  the  left 
to  the  right,  forming  the  parallelopiped  L  V  T  R  M  K  P. 
At  the  same  time  the  triangular  prisms  B  II  D  1 
and  B  C  D  E  A  N  will  be  changed  into  the  prisms  VPB 
M   K  L.  and  V  T  B  M  L  Q. 

If  possible,  follow  the  directions  just  given  with  a  block 
of  wood  or  a  piece  of  potato  in  the  proper  shape. 

Son  many  faces  of  the  new  parallelopiped  will  be  rec- 
tangles ?  If  the  new  parallelopiped  is  imagined  to  stand  on 
tin*  bast*  V  T  l\  l\  it  will  be  I  right  parallelopiped,  beCftOM 
its  edges,  K  1\  M  It,  etc.,  are  at  right  angles  tc  the  two 
faces.  Illustrate  this  by  showing  that  your  block,  if  placed 
on  the  base  V  T  R  P,  or  M  K  L  Q,  will  stand  upright,  bat 
that  if  it  is  placed  on  any  other  base  it  will  lean  over.  The 
volume  of  the  right  parallelopiped,  by  639,  will  be  the 
product  of  tin  T  |{  V  bv  the  altitude  K  P.     The 

triangular  prisms  V  P  B  M  K  L.  and  V  T  B  M  L  Q,  are 
right  prisms.  (Why  ?)  The  volume  of  one  of  them  is,  by 
642,  the  product  of  the  bow  V   P  B  and  the  altitude  K  I\ 


CONSTRUCTIONAL   GEOMETRY  129 

or  just  one  half  of  the  volume  of  the  right  parallelopiped. 
Hence  a  right  parallelopiped,  as  well  as  a  rectangular  paral- 
lelopiped (640),  can  be  cut  into  two  parts  which  have  the 
same  volume  by  a  plane  through  two  diagonally  opposite 
edges.  But  the  right  parallelopiped  of  Fig.  99  occupies  the 
same  space  as  the  original  oblique  parallelopiped  ;  and  the 
right  triangular  prisms  occupy  the  same  space  as  the  oblique 
triangular  prisms,  so  that  any  parallelopiped— whether  rec- 
tangular, right,  or  oblique— can  be  cut  into  two  parts  which 
have  the  same  volume  by  a  plane  through  two  diagonally  op- 
posite edges. 

644.  Test  the  principle  just  stated  in  643  by  weighing 
the  two  pieces  into  which  parallelopipeds  of  wood  having  the 
required  shapes  are  cut  by  planes  containing  two  diagonally 
opposite  edges. 

645.  The  volume  of  the  oblique  parallelopiped 
ABCDEFH  (Fig.  99)  is,  by  639,  the  product  of  its  base 
13  C  D  II  by  its  altitude,  the  perpendicular  distance  between 
its  two  bases ;  the  volume  of  the  oblique  prism  B  D  II 
E  F  A,  which  is  one  half  of  the  parallelopiped  by  643,  is 
the  product  of  one  half  of  B  0  D  II,  or  13  D  H,  by  its  alti- 
tude, that  is,  it  is  the  product  of  its  base  B  D  II  by  its  alti- 
tude, so  that  the  volume  of  any  triangular  prism  is  the  prod- 
uct of  its  base  by  its  altitude. 

646.  In  how  many  different  ways  could  you  cut  an  ob- 
lique parallelopiped  into  triangular  prisms  by  planes  con- 
taining two  diagonally  opposite  edges  ?  Would  there  be 
any  difference  in  size  in  the  various  prisms  ?  in  shape  ? 

647.  Make  three  lines,  a,  b,  and  c  ;  form  a  right  parallel- 
opiped equivalent  to  the  product  of  the  lines  a  x  b  x  c ; 
also  form  an  oblique  parallelopiped  equivalent  to  a  x  b  x  c  ; 
form  an  oblique  triangular  prism  equivalent  to  a  x  b  x  c. 

648.  A  tin  box,  in  form  a  right  triangular  prism,  has  as 
its  bases  two  right-angled  isosceles  triangles  with  the  equal 
sides  8  cm.  long  ;  the  box  is  15  cm.  tall.  How  many  ccm. 
of  water  will  the  box  hold  ?  How  many  sq.  cm.  of  tin  will 
be  needed  for  the  box  and  its  cover  ? 

9 


130  ELEMENTARY  AND 

649.  The  inside  dimensions  of  a  cistern  shaped  like  a 
rectangular  parallelopiped  are  10  ft.  x  6  f t.  x  8  ft.  ;  how 
many  gallons  of  water  will  the  cistern  hold,  if  it  takes  7i 
gallons  to  fill  a  cubic  foot  of  space  ?• 

650.  A  right  prism  10  in.  tall  has  a  regular  hexagon  for  a 
base  ;  if  each  edge  of  the  base  is  6  in.,  how  much  space 
does  the  prism  occupy  ? 

651.  A  water-pail  in  the  form  of  a  cylinder  is  12  in.  tall 
and  9  in.  in  diameter;  can  you  tell  how  many  gallons  ii 
will  hold  if  it  takes  231  cu.  in.  to  hold  a  gallon  ?  (The 
volume  of  a  cylinder  is  found  on  the  same  principle  on 
which  the  volume  of  a  prism  is  found.) 

652.  Into  a  tumbler  partly  filled  with  water  put  a  piece 
of  lead  or  a  block  of  wood  of  irregular  shape,  and  note 
carefully  how  much  tin-  water  rises  when  the  object  is  com- 
pletely under  water.  Estimate  the  volume  of  the  im- 
mersed solid  after  taking  the  proper  measurements.  Use 
a  tumbler  that  has  the  form  of  a  cylinder. 

.  653.  Sink  in  a  tumbler  of  water  a  triangular  prism,  and 
note  the  height  to  which  the  water  rises;  next,  sink  a  tri- 
angular pyramid  with  a  base  and  an  altitude  equal  to  those 
respectively  of  the  prism  ;  how  much  more  space  does  the 
prism  occupy  than  the  pyramid  does  t  Try  this  experi- 
ment several  times  before  giving  a  final  answer,  and,  if  pos- 
sible, use  several  pairs  of  prisms  and  pyramids,  making  each 
pair  according  to  the  directions  above,  but  varying  the 
sizes  of  the  blocks  used. 

»'..".  1.  Form  two  pyramids  of  the  same  height  on  triangu- 
lar bases  that  have  the  same  area  hut  very  different  forms. 
Sink  these  pyramids  in  water,  and  note  whether  one  occu- 
pies more  space  than  the  other. 

655.  Form  several  triangular  pyramids  of  the  same  altitude 
and  with  equivalent  (not  equal)  bases,  and  form  one  prism 
with  an  altitude  equal  to  that  of  the  pyramids,  and  with  a 
base  equivalent  to  one  of  the  bases  of  the  pyramid.  See 
how  many  pyramids  it  takes  to  displace  as  much  water  as 
the  prism  displaces. 


CONSTRUCTIONAL   GEOMETRY  131 

656.  Sum  up  the  results  of  the  experiments  in  653-655 
by  writing  a  principle  that  will  enable  you  to  obtain  the 
volume  of  a  triangular  pyramid. 

657.  Can  you  extend  the  principle  of  656  so  that  it  will 
include  any  pyramid  and  also  a  cone  ? 

658.  In  dealing  with  triangles  you  saw  that  you  could 
move  the  vertex  of  a  triangle  along  a  line  parallel  to  the 
base  without  altering  the  area  of  the  triangle.  Do  your  ex- 
periments in  653-655  point  to  any  similar  principle  about 
triangular  pyramids  ?     What  is  the  principle  ? 

In  moving  the  vertex  of  a  triangle  without  changing  the 
area  you  were  obliged  to  move  it  either  symparallel  with 
the  base  or  antiparallel  to  it.  But  in  moving  the  vertex 
of  a  triangular  pyramid,  you  can  move  it  in  a  countless 
number  of  directions  without  changing  the  volume  of  the 
pyramid  ;  for  a  line  will  be  parallel  to  the  triangular  base 
of  the  pyramid  if  it  is  parallel  to  any  one  of  the  countless 
lines  that  can  be  imagined  in  the  triangular  base.  To  get 
a  clear  idea  of  this,  place  a  triangular  pyramid  on  the  table ; 
you  will  at  once  see  that  there  are  thousands  of  directions 
in  which  you  could  move  the  vertex  without  changing  the 
volume,  keeping  it  in  a  line  parallel  to  the  base  or  the 
table.  You  can  therefore 
mould  triangular  pyramids 
with  even  more  freedom  than 
you  can  mould  triangles. 

659.  By  moulding  triangles 
you  can  change  a  polygon 
into  an  equivalent  triangle. 
In  the  same  way,  by  moulding 
triangular  pyramids  you  can 
change  a  pyramid  which  has  a 
polygon  as  a  base  into  a  trian- 
gular pyramid.  For  an  exam- 
ple see  Fig.  100,  where  the  ¥m-  m 
pyramid  V— A  BOD  has  a  quadrilateral  as  a  base. 

The  plane  V  B  D  cuts  the  pyramid  into  two  triangular 


132  ELEMENTARY  AND 

pyramids,  V-A  B  D  and  V-B  C  D.  The  latter  may  be 
imagined  to  stand  on  the  base  B  D  V,  with  vertex  at  C. 
Move  C  along  the  line  C  K,  which  is  parallel  to  D  B,  and 
therefore  to  V  B  D.  (See  658.)  The  pyramid  C-BDV 
is  thus  moulded  into  the  equivalent  pyramid  K-BOV, 
which  joins  with  V— A  B  D  to  form  the  triangular  pyra- 
mid V  — A  K  I) ;  equivalent  to  the  original  pyramid. 

600.  Draw  three  different  pyramids  with  quadrilaterals 
for  bases,  and  mould  each  pyramid  into  its  equivalent  tri- 
angular pyramid. 

001.  Can  you  mould  two  triangular  pyramids  of  the  same 
height,  but  with  different  bases,  into  one  triangular  pyra- 
mid ? 

Review  the  method  by  which  you  lowered  the  ver- 
tex of  a  triangle  to  any  desired  point  without  changing  the 
surface  of  the  triangle,  and  then  study  the  construction  which 
follows,  by  which  the  vortex  of  a  pyramid  is  lowered  with- 
out anv  change  in  the  volume  of  the  pyramid.  The  origi- 
nal pyramid  is  V— A  B  C,  Fig.  101,  and  the  point  to 
which  the  yertex  is  to  be  lowered  ii  P. 

My  a  plane,  P  B  0  (a  guide  plane  corresponding  to  the 
guide  line  of  Exercise  419),  a  pyramid,  V  — P  BC,  is  cut  off; 
the  vertex  V  is  then  moved  to  K 
along  V  K,  which  is  parallel  to 
P  B,  and  therefore  parallel  to  the 
base  I1  H  ('.  which  contains  P  B  ; 
the  edges  P  V,  B  V,  C  V,  take  the 
respective  positions,  P  K,  B  K . 
c  K.  and  the  pyramid  V-A  BC 
is  changed  Into  the  equivalent 
pyramid  P— A  K  <\ 

668.  Draw  two  triangular  pyra- 
mids with  bases  on  the  same  plane, 
but  with  di  tie  rent  altitudes.  Low- 
er the  taller  pyramid  to  the  level 
of  the  shorter  one.  Then  find  one  triangular  pyramid  equiv- 
alent to  their  sum. 


CONSTRUCTIONAL   GEOMETRY  133 

664.  Can  you  raise  the  vertex  of  a  triangular  pyramid 
to  any  desired  level  without  changing  the  volume  of  the 
pyramid  ? 

665.  How  much  larger  is  a  triangular  prism  than  a  tri- 
angular pyramid  which  stands  on  the  same  base  and  has 
the  same  altitude  as  the  prism  ?    (See  656.) 

666.  If  you  wish  to  find  a  triangular  pyramid  with  a 
volume  equal  to  that  of  a  given  triangular  prism,  how  tall 
must  you  make  the  pyramid,  if  you  make  its  base  equal  to 
the  base  of  the  prism  ?  How  large  a  base  must  it  have, 
if  you  keep  the  height  the  same  as  that  of  the  prism  ? 

667.  A  wine-glass  shaped  like  a  cone  and  a  tumbler 
shaped  like  a  cylinder  are  of  the  same  height  and  have 
the  same  width  at  the  top  ;  how  much  more  water  will  the 
tumbler  hold  than  the  wine-glass  ? 

668.  How  much  more  than  the  wine-glass  would  the  tum- 
bler of  667  hold,  if  it  were  twice  as  tall  as  the  wine-glass, 
other  conditions  remaining  the  same  ? 

669.  In.  432  and  the  following  articles  you  saw  that  a 
circle  could  be  moulded  into  a  triangle  with  the  radius 
as  altitude,  if  only  a  line  could  be  found  for  the  base  equal 
in  length  to  the  circumference,  and  by  an  experiment 
you  found  how  to  get  the  circumference  from  the  radius 
with  sufficient  accuracy  for  practical  purposes.  In  a  simi- 
lar way  a  sphere  can  be  moulded  into  a  cone  with  an  altitude 
equal  to  the  radius,  if  it  is  possible  to  find  a  base  equiva- 
lent to  the  surface  of  the  sphere.  The  following  experi- 
ment will  fix  in  your  minds  the  facts  that  have  been  dis- 
covered about  the  surface  of  a  sphere  by  more  advanced 
principles  of  Geometry  than  you  have  yet  taken  up.  Take 
a  ball  of  wood  or  of  rubber  (a  baseball  would  answer  very 
well)  and  have  a  tin  cylindrical  box  made,  into  which  the 
ball  will  just  fit.  When  the  cover  is  on  the  box  it  should 
touch  the  ball.  Sink  first  the  ball  and  then  the  closed  box 
in  a  small  cylindrical  jar  of  water,  and  notice  how  much 
higher  the  water  rises  in  one  case  than  in  the  other.  You 
will  find  that  the  ball  makes  the  water  rise  only  two-thirds 


13±  ELEMENTARY  AND 

as  far  as  the  box  does  ;  so  that  the  ball  occupies  only  two- 
thirds  as  much  space  as  the  cylinder  which  just  surrounds 
it.  Now  imagine  the  ball  moulded  into  a  cone,  with  the 
radius  of  the  ball  as  its  altitude,  and  with  the  surface 
of  the  ball  flattened  into  a  circle  for  its  base.  The 
cone  will  be  only  one-half  as  tall  as  the  box,  for  the  box 
has  the  diameter  of  the  ball  for  its  altitude,  and  the  cone 
has  the  radius  of  the  ball  for  its  altitude.  Since  the  coue 
is  one-half  as  tall  as  the  box,  to  occupy  as  much  space  as 
thr  box  does  it  would  need  a  base  six  times  as  large  as  the 
base  of  the  box  (see  668) ;  but  since  it  occupies  only  two- 
thirds  as  much  space  as  the  box  does,  its  base  need  be  only 
t\\  .-thirds  of  six  times  as  large  as  the  base  of  the  box,  or 
four  times  the  base  of  the  box.  This  experiment  illustrates 
the  truth,  which  can  be  proved  by  the  principles  of  Geom- 
etry, that  the  surface  of  a  sphere  is  four  times  the  base 
of  a  cylinder  in  which  the  sphere  is  inscribed. 

670.  If  you  should  cut  the  bull  in  halves,  and  then  re- 
place the  hal\<<  in  the  lx>x.  the  flat  side  of  one  half  would 
exactly  fit  the  bottom  of  the  box,  and  the  flat  side  of  the 
other  half  would  lit  the  top  of  the  box,  while  thecur\.«l 
portions  would  just  touch  each  other  in  the  centre  of  the 
box.    What  portion  <»f  the  box  is  empty  P 

If  you  cut  an  orange  in  halves,  the  two  circles  forming 
the  plane  sides  of  the  pieces  contain  as  much  surface  as  the 
peel  (.n  OIM  Of  the  pieces.      Why  P 

671.  If  yon  should  cut  the  box  used  in  o'*.*.»  and  670 along 
a  line  perpendicular  to  the  two  bases,  and  flatten  the  lateral 
surface  into  a  plane  surface,  you  would  form  a  rectangle 
with  the  ci  renin  ference  of  the  original  base  as  abase,  and 
with  an  altitude  equal  to  twice  the  radius  of  the  base.  If 
you  should  also  mould  one  of  the  bases  into  a  triangle,  the 
triangle  would  have  the  circumference  of  the  original  base 
as  a  base,  and  an  altitude  equal  to  the  radios.  Since  the 
rectangle  has  the  same  base  as  the  triangle,  but  twice  its 
altitude,  the  surface  of  the  rectangle  is  four  times  that  of 
the  triangle.     Make  this  clear  by  drawing  a  rectangle  and 


CONSTRUCTIONAL   GEOMETRY  135 

a  triangle  which  fulfil  the  above  conditions,  and  by  chang- 
ing the  triangle  into  a  rectangle.  Since  the  lateral  surface 
of  the  cylindrical  box  is  four  times  that  of  its  base,  it  must 
be  equivalent  to  that  of  the  ball.  Hence  another  way  of 
describing  the  surface  of  a  sphere  :  The  surface  of  a  sphere 
is  the  same  as  the  lateral  surface  of  a  cylinder  in  which  the 
sphere  is  inscribed. 

672.  Compare  the  entire  surface  of  a  cylinder  with  that 
of  a  sphere  which  is  inscribed  in  the  cylinder. 

673.  Can  you  draw  a  circle  that  shall  have  as  much  sur- 
face as  a  given  sphere  has  ?  A  sphere  is  given  if  its  radius 
is  known. 

674.  Can  you  draw  a  cone  that  shall  have  a  base  whose 
radius  is  the  radius  of  a  given  sphere,  and  that  at  the  same 
time  shall  equal  the  sphere  in  volume  ? 

675.  Find  by  arithmetic  the  volume  of  each  of  the  fol- 
lowing spheres,  the  radii  of  which  are  7  in.,  6  in.,  3  in., 
and  14  in.,  respectively.  Compare  the  first  answer  with 
the  last  answer,  and  the  second  answer  with  the  third  an- 
swer. 

676.  Assuming  that  the  earth  is  a  sphere  with  a  radius  of 
4,000  miles,  how  many  square  miles  are  there  on  the  earth's 
surface  ? 

677.  As  a  result  of  your  study  of  this  section  write  and 
record  a  rule  for  finding  the  volume  of  each  of  the  following 
solids  :  a  parallelopiped  ;  a  prism  ;  a  pyramid  ;  a  cone  ;  a 
cylinder  ;  and  a  sphere. 

678.  Draw  three  lines,  a,  b,  and  c,  and  form  a  parallelo- 
piped, a  prism,  a  triangular  pyramid,  and  a  quadrangular 
pyramid,  equivalent  to  the  product  of  these  lines.  Can 
you  draw  a  cone,  a  cylinder,  and  a  sphere  that  shall  be 
equivalent  to  the  product  of  these  lines  ? 


INDEX 


Altitude,  242. 

Angle,  76;  Right,  Acute,  Obtuse, 
Oblique,  Straight  Angles,  77; 
Exterior,  Interior,  Alternate 
Exterior  or  Interior,  Corre- 
sponding Angles,  121;  Inscribed 
Angle,  569. 

Antiparallel,  113. 

Arc,  566. 

Area,  371. 

Axiom,  61. 

Chord,  569. 
Circumference,  32. 
Contact,  Point  of,  588. 
Corollary,  213. 

Diagonal,  297. 
Dimensions,  4. 
Dividing  Tool,  355. 
Drawing  to  Scale,  183. 

Edges,  12. 
Ellipse,  38. 
Equilateral,  219. 
Equivalent,  373. 

Faces,  12. 

Geometrical  Solids,  4. 

Hexagon,  404. 
Hypotenuse,  272. 


Inscribe,  312. 
Inscribed  Angle,  569. 
Interior  Tangent,  624. 
Isosceles,  210. 

Lines,  12,  26,  and  27. 

Median  of  a  Trapezoid,  333. 

Parallel,  113. 

Parallelogram,  148. 

Parallelopiped,  627,  632,  643. 

Pentagon,  97 

Perimeter,  379. 

Perpendicular,  224 

Points,  17. 

Polygon,  428. 

Prism,  640. 

Product  of  Two  Lines,  368;    of 

Three  Lines,  629. 
Proof,  150. 

Proportional  Dividers,  351. 
Protractor,  83. 
Pythagorean  Principle,  463. 

Quadrilateral,  100. 

Radius,  33. 

Rectangle,  290. 

Rectangular  Parallelopiped,  627. 

Rhomboid,  290. . 

Rhombus,  290. 


138 


INDEX 


Right   Angle,   77;    Right   Prism,       Stood  on,  574. 


640;  Right  Parallelopiped,  643. 
Root,  508. 


Scalene,  324. 
Scholium,  121. 
Segment,  582. 
Similar,  349. 
Solid*,  3,  4,  58. 
Square,  290. 
Square  Root,  508. 


Straight  Angle,  79. 
Supplement,  119. 
Surface,  12,  47. 
Symparallel,  113. 

Tangent,  588,  624. 
Trapezium,  2H1. 
Trapezoid,  281. 
Triangles,  105,  210,  324. 

Vebtex,  87. 


8ION8   USED   IN    ADDITION  TO   TIIE    ORDINARY    8I0N8    OP 
ARITHMII  l«  . 


= ,  it  equal  to. 
4  •  is  not  equal  to. 
<•  is  smaller  than. 
>  *  is  larger  than. 
/_>   /k*  angle,  angles 

A*  ^.  Mugte,  triangles. 
|.  is  parallel  to 

1 1  .  is  symparallel  with. 

f  I  .  is  antiparallel  to. 
_L  »  is  perptuditular  to. 


■O",  is  equivalent  to. 
.  ' ..  therefore. 
1         L  rectangle. 
Ji 7  •  rarallelogram,  or  rhom- 
boid. 

Oi   ®,  eireuniferem  e.     <  ire inii- 

ferences. 
O.  €>.  eirele,  circles. 
*""\  arc. 
*J      ,  square  root. 


.1JB 

Or 


-"S* 


or  r 

J£ORH\h. 


SUMMAKY  OF 
FACTS,  DEFINITIONS,  AND  PRINCIPLES 


SUMMAliV    OF   FACTS,    I  >i:iIM  TI<  >NS,   AND    PRDfCD 

N-.iK.  —  Each  pupil  should  enter  in  thin  Summary  tin-  important 
principle-  and  definition*,  when  he  learns  them,  expressing  them  in  tin- 
language  that  MOM  most  natural  to  himself,  provided  that  his  words 
accurately  express  the  ideas  intended. 

An  index  should  be  made,  at  the  time  of  the  entries,  on  the  last  Hank 

paf* 

Kadi  teacher  will  hare  his  own  opinion  of  the  principles  that  he 
wishes  his  particular  class  to  become  familiar  with ;  so  that  the  entries 
Ik  -re  ought  to  vary  with  different  teachers,  and  also  with  different 
classes  under  the  same  teacher 

To  assist  those  who  have  no  wish  to  make  a  selection  for  them- 
it  is  suggested  that  entries  ought  to  be  made,  in  general,  after  tli 
of  excr  .  76,   77.   7ft,   IIS, 

111,  11'.,  11H,  120,  121.  144.  14*.»,  150,  156,  L60,  161,  166,  17-.  l.'L'.  I '.»;;, 
106,  111,  IIS,  118,  114)  Mft,  MS,  272,  278,  297,  809,  31H,  319,  M 
::;:;:.  :::i.  H  ■  "-71.  581, 

658,  669,  671,  and  077. 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.00  ON  THE  SEVENTH  DAY 
OVERDUE. 


OCT  11 1937  v... 

SjMPW 

AUG  29 1963 

mil 

rftc  v  L-f 

0CT2  3'63-9AM 

JUL  231992 

3v5!  it 

utobiSCCIRC  OCT ; 

.4'93 

LD  21-95m  7,37 

YB  05206 


III 


U.C.  BERKELEY  LIBRARIES 


C0312MAQ7E 


183979 


